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NCERT Solutions for Exercise 7.5 Class 12 Maths Chapter 7 - Integrals

NCERT Solutions for Exercise 7.5 Class 12 Maths Chapter 7 - Integrals

Edited By Ramraj Saini | Updated on Dec 03, 2023 09:57 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.5

NCERT Solutions for Exercise 7.5 Class 12 Maths Chapter 7 Integrals are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for Class 12 Maths chapter 7 exercise 7.5 deals with some of the functions which are not discussed yet in earlier exercises. It includes rational functions and advanced levels of logarithmic functions. If students practice this NCERT book exercise diligently, they can attain a good level of understanding of Integrals. Exercise 7.5 Class 12 Maths questions can be seen verbatim in CBSE board examinations. NCERT solutions for Class 12 Maths chapter 7 exercise 7.5 along with some in text examples is recommended to be solved .

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12th class Maths exercise 7.5 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Integrals Class 12 Chapter 7 Exercise: 7.5

Question:1 Integrate the rational functions \frac{x }{( x +1)( x+2)}

Answer:

Given function \frac{x }{( x +1)( x+2)}

Partial function of this function:

\frac{x }{( x +1)( x+2)} = \frac{A}{(x+1)}+\frac{B}{(x+2)}

\implies x = A(x+2)+B(x+1)

Now, equating the coefficients of x and constant term, we obtain

A+B =1

2A+B =0

On solving, we get

A=-1\ and\ B =2

\therefore \frac{x}{(x+1)(x+2)} = \frac{-1}{(x+1)}+\frac{2}{(x+2)}

\implies \int \frac{x}{(x+1)(x+2)} dx =\int \frac{-1}{(x+1)}+\frac{2}{(x+2)} dx

=-\log|x+1| +2\log|x+2| +C

=\log(x+2)^2-\log|x+1|+C

=\log\frac{(x+2)^2}{(x+1)}+C

Question:2 Integrate the rational functions \frac{1}{x^2 -9 }

Answer:

Given function \frac{1}{x^2 -9 }

The partial function of this function:

\frac{1}{(x+3)(x-3)}= \frac{A}{(x+3)}+\frac{B}{(x-3)}

1 = A(x-3)+B(x+3)

Now, equating the coefficients of x and constant term, we obtain

A+B =1

-3A+3B =1

On solving, we get

A=-\frac{1}{6}\ and\ B =\frac{1}{6}

\frac{1}{(x+3)(x-3)}= \frac{-1}{6(x+3)} +\frac{1}{6(x-3)}

\int \frac{1}{(x^2-9)}dx = \int \left ( \frac{-1}{6(x+3)}+\frac{1}{6(x-3)} \right )dx

=-\frac{1}{6}\log|x+3| +\frac{1}{6}\log|x-3| +C

= \frac{1}{6}\log\left | \frac{x-3}{x+3} \right |+C

Question:3 Integrate the rational functions \frac{3x -1}{( x-1)(x-2)(x-3)}

Answer:

Given function \frac{3x -1}{( x-1)(x-2)(x-3)}

Partial function of this function:

\frac{3x -1}{( x-1)(x-2)(x-3)}= \frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}

3x-1 = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) .(1)

Now, substituting x=1,2,\ and\ 3 respectively in equation (1), we get

A =1,\ B=-5,\ and\ C=4

\therefore \frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{1}{(x-1)} -\frac{5}{(x-2)}+\frac{4}{(x-3)}

That implies \int \frac{3x-1}{(x-1)(x-2)(x-3)} dx = \int \left \{ \frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)} \right \}dx

= \log|x-1|-5\log|x-2|+4\log|x-3|+C

Question:4 Integrate the rational functions \frac{x }{( x-1)(x-2)(x-3)}

Answer:

Given function \frac{x }{( x-1)(x-2)(x-3)}

Partial function of this function:

\frac{x }{( x-1)(x-2)(x-3)}= \frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}

x = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) .....(1)

Now, substituting x=1,2,\ and\ 3 respectively in equation (1), we get

A =\frac{1}{2},\ B=-2,\ and\ C=\frac{3}{2}

\therefore \frac{x}{(x-1)(x-2)(x-3)} = \frac{1}{2(x-1)} -\frac{2}{(x-2)}+\frac{3}{2(x-3)}

That implies \int \frac{x}{(x-1)(x-2)(x-3)} dx = \int \left \{ \frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)} \right \}dx

= \frac{1}{2}\log|x-1|-2\log|x-2|+\frac{3}{2}\log|x-3|+C

Question:5 Integrate the rational functions \frac{2x}{x^2 + 3x +2 }

Answer:

Given function \frac{2x}{x^2 + 3x +2 }

Partial function of this function:

\frac{2x}{x^2 + 3x +2 }= \frac{A}{(x+1)}+\frac{B}{(x+2)}

2x = A(x+2)+B(x+1) ...........(1)

Now, substituting x=-1\ and\ -2 respectively in equation (1), we get

A ={-2},\ B=4

\frac{2x}{x^2 + 3x +2 }= \frac{-2}{(x+1)}+\frac{4}{(x+2)}

That implies \int \frac{2x}{x^2 + 3x +2 }dx= \int \left \{ \frac{-2}{(x+1)}+\frac{4}{(x+2)} \right \}dx

=4\log|x+2| -2\log|x+1| +C

Question:6 Integrate the rational functions \frac{1- x^2 }{ x ( 1- 2x )}

Answer:

Given function \frac{1- x^2 }{ x ( 1- 2x )}

Integral is not a proper fraction so,

Therefore, on dividing (1-x^2) by x(1-2x) , we get

\frac{1- x^2 }{ x ( 1- 2x )} = \frac{1}{2} +\frac{1}{2}\left ( \frac{2-x}{x(1-2x)} \right )

Partial function of this function:

\frac{2-x}{x(1-2x)} =\frac{A}{x}+\frac{B}{(1-2x)}

(2-x) =A(1-2x)+Bx ...........(1)

Now, substituting x=0\ and\ \frac{1}{2} respectively in equation (1), we get

A =2,\ B=3

\therefore \frac{2-x}{x(1-2x)} = \frac{2}{x}+\frac{3}{1-2x}

No, substituting in equation (1) we get

\frac{1-x^2}{(1-2x)} = \frac{1}{2}+\frac{1}{2}\left \{ \frac{2}{3}+\frac{3}{(1-2x)} \right \}

\implies \int \frac{1-x^2}{x(1-2x)}dx =\int \left \{ \frac{1}{2}+\frac{1}{2}\left ( \frac{2}{x}+\frac{3}{1-2x} \right ) \right \}dx

=\frac{x}{2}+\log|x| +\frac{3}{2(-2)}\log|1-2x| +C

=\frac{x}{2}+\log|x| -\frac{3}{4}\log|1-2x| +C

Question:7 Integrate the rational functions \frac{x }{( x^2+1 )( x-1)}

Answer:

Given function \frac{x }{( x^2+1 )( x-1)}

Partial function of this function:

\frac{x }{( x^2+1 )( x-1)} = \frac{Ax+b}{(x^2+1)} +\frac{C}{(x-1)}

x = (Ax+B)(x-1)+C(x^2+1)

x=Ax^-Ax+Bc-B+Cx^2+C

Now, equating the coefficients of x^2, x and the constant term, we get

A+C = 0

-A+B =1 and -B+C = 0

On solving these equations, we get

A = -\frac{1}{2}, B= \frac{1}{2},\ and\ C=\frac{1}{2}

From equation (1), we get

\therefore \frac{x}{(x^2+1)(x-1)} = \frac{\left ( -\frac{1}{2}x+\frac{1}{2} \right )}{x^2+1}+\frac{\frac{1}{2}}{(x-1)}

\implies \int \frac{x}{(x^2+1)(x-1)}

=-\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx+\frac{1}{2} \int \frac{1}{x-1}dx

=- \frac{1}{4} \int \frac{2x}{x^2+1} dx +\frac{1}{2} \tan^{-1}x + \frac{1}{2} \log|x-1| +C

Now, consider \int \frac{2x}{x^2+1} dx ,

and we will assume (x^2+1) = t \Rightarrow 2xdx =dt

So, \int \frac{2x}{x^2+1}dx = \int \frac{dt}{t} =\log|t| = \log|x^2+1|

\therefore \int \frac{x}{(x^2+1)(x-1)} =-\frac{1}{4}\log|x^2+1| +\frac{1}{2}\tan^{-1}x +\frac{1}{2}\log|x-1| +C or

\frac{1}{2}\log|x-1| - \frac{1}{4}\log|x^2+1|+\frac{1}{2}\tan^{-1}x +C

Question:8 Integrate the rational functions \frac{x }{( x+1)^2 ( x+2)}

Answer:

Given function \frac{x }{( x+1)^2 ( x+2)}

Partial function of this function:

\frac{x }{( x+1)^2 ( x+2)} = \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+2)}

x = A(x-1)(x+2)+B(x+2)+C(x-1)^2

Now, putting x=1 in the above equation, we get

B =\frac{1}{3}

By equating the coefficients of x^2 and constant term, we get

A+C=0

-2A+2B+C = 0

then after solving, we get

A= \frac{2}{9}\ and\ C=\frac{-2}{9}

Therefore,

\frac{x}{(x-1)^2(x+2)} = \frac{2}{9(x-1)}+\frac{1}{3(x-1)^2}-\frac{2}{9(x+2)}

\int \frac{x}{(x-1)^2(x+2)}dx= \frac{2}{9}\int \frac{1}{(x-1)}dx+\frac{1}{3}\int \frac{1}{(x-1)^2}dx-\frac{2}{9}\int \frac{1}{(x+2)}dx

= \frac{2}{9}\log|x-1|+\frac{1}{3}\left ( \frac{-1}{x-1} \right )-\frac{2}{9}\log|x+2|+C

\frac{2}{9}\log\left | \frac{x-1}{x+2} \right | -\frac{1}{3(x-1)}+C

Question:9 Integrate the rational functions \frac{3x+ 5 }{x^3 - x^2 - x +1 }

Answer:

Given function \frac{3x+ 5 }{x^3 - x^2 - x +1 }

can be rewritten as \frac{3x+ 5 }{x^3 - x^2 - x +1 } = \frac{3x+5}{(x-1)^2(x+1)}

Partial function of this function:

\frac{3x+5}{(x-1)^2(x+1)}= \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}

3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)^2

3x+5 = A(x^2-1)+B(x+1)+C(x^2+1-2x) ................(1)

Now, putting x=1 in the above equation, we get

B =4

By equating the coefficients of x^2 and x , we get

A+C=0

B-2C =3

then after solving, we get

A= -\frac{1}{2}\ and\ C=\frac{1}{2}

Therefore,

\frac{3x+5}{(x-1)^2(x+1)}= \frac{-1}{2(x-1)}+\frac{4}{(x-1)^2}+\frac{1}{2(x+1)}

\int \frac{3x+5}{(x-1)^2(x+1)}dx= \frac{-1}{2}\int \frac{1}{(x-1)}dx+4\int \frac{1}{(x-1)^2} dx+\frac{1}{2}\int \frac{1}{(x+1)}dx

= -\frac{1}{2}\log|x-1| +4\left ( \frac{-1}{x-1} \right ) +\frac{1}{2}\log|x+1| +C

=\frac{1}{2}\log|\frac{x+1}{x-1}| - \frac{4}{(x-1)} + +C

Question:10 Integrate the rational functions \frac{2x -3 }{(x^2 -1 )( 2x+3)}

Answer:

Given function \frac{2x -3 }{(x^2 -1 )( 2x+3)}

can be rewritten as \frac{2x -3 }{(x^2 -1 )( 2x+3)} = \frac{2x-3}{(x+1)(x-1)(2x+3)}

The partial function of this function:

\frac{2x -3 }{(x^2 -1 )( 2x-3)} = \frac{A}{(x+1)} +\frac{B}{(x-1)}+\frac{C}{(2x+3)}

\Rightarrow (2x-3) =A(x-1)(2x+3)+B(x+1)(2x+3)+C(x+1)(x-1) \Rightarrow (2x-3) =A(2x^2+x-3)+B(2x^2+5x+3)+C(x^2-1) \Rightarrow (2x-3) =(2A+2B+C)x^2+(A+5B)x+(-3A+3B-C)

Equating the coefficients of x^2\ and\ x , we get

B=-\frac{1}{10},\ A =\frac{5}{2},\ and\ C= -\frac{24}{5}

Therefore,

\frac{2x -3 }{(x^2 -1 )( 2x-3)} = \frac{5}{2(x+1)} -\frac{1}{10(x-1)}-\frac{24}{5(2x+3)}

\implies \int \frac{2x-3}{(x^2-1)(2x+3)}dx = \frac{5}{2}\int \frac{1}{(x+1)}dx -\frac{1}{10}\int \frac{1}{x-1}dx -\frac{24}{5}\int \frac{1}{(2x+3)}dx = \frac{5}{2}\log|x+1| -\frac{1}{10}\log|x-1| -\frac{24}{10}\log|2x+3|

= \frac{5}{2}\log|x+1| -\frac{1}{10}\log|x-1| -\frac{12}{5}\log|2x+3|+C

= -\frac{1}{2}\log|x-1| +4\left ( \frac{-1}{x-1} \right ) +\frac{1}{2}\log|x+1| +C

=\frac{1}{2}\log|\frac{x+1}{x-1}| - \frac{4}{(x-1)} + +C

= \frac{2}{9}\log|x-1|+\frac{1}{3}\left ( \frac{-1}{x-1} \right )-\frac{2}{9}\log|x+2|+C

\frac{2}{9}\log\left | \frac{x-1}{x+2} \right | -\frac{1}{3(x-1)}+C

Question:11 Integrate the rational functions \frac{5x}{(x+1)(x^2-4)}

Answer:

Given function \frac{5x}{(x+1)(x^2-4)}

can be rewritten as \frac{5x}{(x+1)(x^2-4)} = \frac{5x}{(x+1)(x+2)(x-2)}

The partial function of this function:

\frac{5x }{(x+1)( x+2)(x-2)} = \frac{A}{(x+1)} +\frac{B}{(x+2)}+\frac{C}{(x-2)}

\Rightarrow (5x) =A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)

Now, substituting the value of x =-1,-2,\ and\ 2 respectively in the equation above, we get

A=\frac{5}{3},\ B =\frac{-5}{2},\ and\ C= \frac{5}{6}

Therefore,

\frac{5x }{(x+1)( x+2)(x-2)} = \frac{5}{3(x+1)} -\frac{5}{2(x+2)}+\frac{5}{6(x-2)}

\implies \int \frac{5x}{(x+1)(x^2-4)}dx = \frac{5}{3}\int \frac{1}{(x+1)}dx -\frac{5}{2}\int \frac{1}{x+2}dx+\frac{5}{6}\int \frac{1}{(x-2)}dx = \frac{5}{3}\log|x+1| -\frac{5}{2}\log|x+2| +\frac{5}{6}\log|x-2|+C

Question:12 Integrate the rational functions \frac{x^3 + x +1}{ x^2-1}

Answer:

Given function \frac{x^3 + x +1}{ x^2-1}

As the given integral is not a proper fraction.

So, we divide (x^3+x+1) by x^2-1 , we get

\frac{x^3 + x +1}{ x^2-1} = x+\frac{2x+1}{x^2-1}

can be rewritten as \frac{2x+1}{x^2-1} =\frac{A}{(x+1)} +\frac{B}{(x-1)}

2x+1 ={A}{(x-1)} +{B}{(x+1)} ....................(1)

Now, substituting x =1\ and\ x=-1 in equation (1), we get

A =\frac{1}{2}\ and\ B=\frac{3}{2}

Therefore,

\frac{x^3+x+1 }{(x^2-1)} =x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)}

\implies \int \frac{x^3+x+1 }{(x^2-1)}dx =\int xdx +\frac{1}{2}\int \frac{1}{(x+1)} dx+\frac{3}{2} \int \frac{1}{(x-1)}dx

= \frac{x^2}{2}+\frac{1}{2}\log|x+1| +\frac{3}{2}\log|x-1| +C

Question:13 Integrate the rational functions \frac{2}{(1-x)(1+ x^2)}

Answer:

Given function \frac{2}{(1-x)(1+ x^2)}

can be rewritten as \frac{2}{(1-x)(1+ x^2)} = \frac{A}{(1-x)}+\frac{Bx+C}{1+x^2}

2 =A(1+x^2)+(Bx+C)(1-x) ....................(1)

2 =A +Ax^2 +Bx-Bx^2+C-Cx

Now, equating the coefficient of x^2, x, and constant term, we get

A-B= 0 , B-C = 0 , and A+C =2

Solving these equations, we get

A=1, B=1,\ and\ C=1

Therefore,

\therefore \frac{2}{(1-x)(1+ x^2)} = \frac{1}{(1-x)}+\frac{x+1}{1+x^2}

\implies \int \frac{2}{(1-x)(1+ x^2)}dx =\int \frac{1}{(1-x)} dx+ \int \frac{x}{1+x^2}dx +\int \frac{1}{1+x^2}dx = -\int \frac{1}{x-1}dx +\frac{1}{2}\int \frac{2x}{1+x^2}dx +\int\frac{1}{1+x^2}dx

=-\log|x-1| +\frac{1}{2}\log|1+x^2| +\tan^{-1}x+C

Question:14 Integrate the rational functions \frac{3x-1}{(x+2)^2}

Answer:

Given function \frac{3x-1}{(x+2)^2}

can be rewritten as \frac{3x-1}{(x+2)^2} = \frac{A}{(x+2)}+\frac{B}{(x+2)^2}

3x-1 = A(x+2)+B

Now, equating the coefficient of x and constant term, we get

A=3 and 2A+B = -1 ,

Solving these equations, we get

B=-7

Therefore,

\frac{3x-1}{(x+2)^2} = \frac{3}{(x+2)}-\frac{7}{(x+2)^2}

\implies \int\frac{3x-1}{(x+2)^2}dx = 3 \int \frac{1}{(x+2)}dx-7\int \frac{x}{(x+2)^2}dx

\implies 3\log|x+2| -7\left ( \frac{-1}{(x+2)}\right )+C

\implies 3\log|x+2| + \frac{7}{(x+2)} +C

Question:15 Integrate the rational functions \frac{1}{x^4 -1 }

Answer:

Given function \frac{1}{x^4 -1 }

can be rewritten as \frac{1}{x^4 -1 } = \frac{1}{(x^2-1)(x^2+1)} =\frac{1}{(x+1)(x-1)(1+x^2)}

The partial fraction of above equation,

\frac{1}{(x+1)(x-1)(1+x^2)} = \frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{Cx+D}{(x^2+1)}

1 = A(x-1)(x^2+1) +B(x+1)(x^2+1)+(Cx+D)(x^2-1)

1 = A(x^3+x-x^2-1)+B(x^3+x+x^2+1)+Cx^3+Dx^2-Cx-D 1 = (A+B+C)x^3 +(-A+B+D)x^2+(A+B-C)x+(-A+B-D)

Now, equating the coefficient of x^3,x^2,x and constant term, we get

A+B+C = 0 and -A+B+D = 0

A+B-C = 0 and -A+B-D = 1

Solving these equations, we get

A= -\frac{1}{4}, B=\frac{1}{4},C=0,\ and\ D = -\frac{1}{2}

Therefore,

\frac{1}{x^4-1} = \frac{-1}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2(x^2+1)}

\implies \int \frac{1}{x^4-1}dx = -\frac{1}{4}\log|x-1| +\frac{1}{4}\log|x-1| -\frac{1}{2}\tan^{-1}x +C

= \frac{1}{4}\log|\frac{x-1}{x+1}| -\frac{1}{2}\tan^{-1}x +C

Question:16 Integrate the rational functions \frac{1}{x ( x^n+1)}

[Hint: multiply numerator and denominator by x ^{n-1} and put x ^n = t ]

Answer:

Given function \frac{1}{x ( x^n+1)}

Applying Hint multiplying numerator and denominator by x^{n-1} and putting x^n =t

\frac{1}{x ( x^n+1)} = \frac{x^{n-1}}{x^{n-1}x(x^n+1)} = \frac{x^{n-1}}{x^n(x^n+1)}

Putting x^n =t

\therefore x^{n-1}dx =dt

can be rewritten as \int \frac{1}{x ( x^n+1)}dx =\int \frac{x^{n-1}}{x^n(x^n+1)}dx = \frac{1}{n} \int \frac{1}{t(t+1)}dt

Partial fraction of above equation,

\frac{1}{t(t+1)} =\frac{A}{t}+\frac{B}{(t+1)}

1 = A(1+t)+Bt ................(1)

Now, substituting t = 0,-1 in equation (1), we get

A=1\ and\ B=-1

\therefore \frac{1}{t(t+1)} = \frac{1}{t}- \frac{1}{(1+t)}

\implies \int \frac{1}{x(x^n+1)}dx = \frac{1}{n} \int \left \{ \frac{1}{t}-\frac{1}{(t+1)} \right \}dx

= \frac{1}{n} \left [ \log|t| -\log|t+1| \right ] +C

= -\frac{1}{n} \left [ \log|x^n| -\log|x^n+1| \right ] +C

= \frac{1}{n} \log|\frac{x^n}{x^n+1}| +C

Question:17 Integrate the rational functions \frac{\cos x }{(1- \sin x )( 2- \sin x )}

[Hint : Put \sin x = t ]

Answer:

Given function \frac{\cos x }{(1- \sin x )( 2- \sin x )}

Applying the given hint: putting \sin x =t

We get, \cos x dx =dt

\therefore \int \frac{\cos x }{(1- \sin x )( 2- \sin x )}dx = \int \frac{dt}{(1-t)(2-t)}

Partial fraction of above equation,

\frac{1}{(1-t)(2-t)} =\frac{A}{(1-t)}+\frac{B}{(2-t)}

1 = A(2-t)+B(1-t) ................(1)

Now, substituting t = 2\ and\ 1 in equation (1), we get

A=1\ and\ B=-1

\therefore \frac{1}{(1-t)(2-t)} = \frac{1}{(1-t)} - \frac{1}{(2-t)}

\implies \int \frac{\cos x }{(1-\sin x)(2-\sin x )}dx = \int \left \{ \frac{1}{1-t}-\frac{1}{(2-t)} \right \}dt

= -\log|1-t| +\log|2-t| +C

= \log\left | \frac{2-t}{1-t} \right |+C

Back substituting the value of t in the above equation, we get

= \log\left | \frac{2-\sin x}{1- \sin x} \right |+C

Question:18 Integrate the rational functions \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )}

Answer:

Given function \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )}

We can rewrite it as: \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} = 1- \frac{(4x^2+10)}{(x^2+3)(x^2+4)}

Partial fraction of above equation,

\frac{(4x^2+10)}{(x^2+3)(x^2+4)} =\frac{Ax+B}{(x^2+3)}+\frac{Cx+D}{(x^2+4)}

4x^2+10 = (Ax+B)(x^2+4)+(Cx+D)(x^2+3)

4x^2+10 = Ax^3+4Ax+Bx^2+4B+Cx^3+3Cx+Dx^2+3D

4x^2+10 = (A+C)x^3+(B+D)x^2+(4A+3C)x+(3D+4B)

Now, equating the coefficients of x^3, x^2, x and constant term, we get

A+C=0 , B+D = 4 , 4A+3C = 0 , 4B+3D =10

After solving these equations, we get

A= 0, B =-2, C=0,\and\ D=6

\therefore \frac{4x^2+10}{(x^2+3)(x^2+4)} = \frac{-2}{(x^2+3)} + \frac{6}{(x^2+4)}

\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} = 1- \left ( \frac{-2}{(x^2+3)} + \frac{6}{(x^2+4)} \right )

\implies \int \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} dx= \int \left \{ 1+ \frac{2}{(x^2+3)} - \frac{6}{(x^2+4)} \right \}dx

= \int \left \{ 1+ \frac{2}{(x^2+(\sqrt3)^2)} - \frac{6}{(x^2+2^2)} \right \}dx

= x+2\left ( \frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt 3} \right ) - 6\left ( \frac{1}{2}\tan^{-1}\frac{x}{2} \right )+C

= x+\frac{2}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3} -3\tan^{-1}\frac{x}{2}+C

Question:19 Integrate the rational functions \frac{2x }{( x^2 +1)( x^2 +3)}

Answer:

Given function \frac{2x }{( x^2 +1)( x^2 +3)}

Taking x^2 = t \Rightarrow 2xdx=dt

\therefore \int \frac{2x }{( x^2 +1)( x^2 +3)}dx = \int \frac{dt}{(t+1)(t+3)}

The partial fraction of above equation,

\frac{1}{(t+3)(t+3)} = \frac{A}{(t+1)}+\frac{B}{(t+3)}

1= A(t+3)+B(t+1) ..............(1)

Now, substituting t = -3\ and\ t = -1 in equation (1), we get

A =\frac{1}{2}\ and\ B = -\frac{1}{2}

\therefore\frac{1}{(t+3)(t+3)} = \frac{1}{2(t+1)}-\frac{1}{2(t+3)}

\implies \int \frac{2x }{( x^2 +1)( x^2 +3)}dx = \int \left \{ \frac{1}{2(t+1)}-\frac{1}{2(t+3)} \right \}dt

= \frac{1}{2}\log|t+1|- \frac{1}{2}\log|t+3| +C

= \frac{1}{2}\log\left | \frac{t+1}{t+3} \right | +C

= \frac{1}{2}\log\left | \frac{x^2+1}{x^2+3} \right | +C

Question:20 Integrate the rational functions \frac{1}{x (x^4 -1)}

Answer:

Given function \frac{1}{x (x^4 -1)}

So, we multiply numerator and denominator by x^3 , to obtain

\frac{1}{x (x^4 -1)} = \frac{x^3}{x^4(x^4-1)}

\therefore \int \frac{1}{x(x^4-1)}dx =\int\frac{x^3}{x^4(x^4-1)}dx

Now, putting x^4 = t

we get, 4x^3dx =dt

Taking x^2 = t \Rightarrow 2xdx=dt

\therefore \int \frac{1}{x(x^4-1)}dx =\frac{1}{4}\int \frac{dt}{t(t-1)}

Partial fraction of above equation,

\frac{1}{t(t-1)} = \frac{A}{t}+\frac{B}{(t-1)}

1= A(t-1)+Bt ..............(1)

Now, substituting t = 0\ and\ t = 1 in equation (1), we get

A = -1\ and\ B=1

\Rightarrow \frac{1}{t(t+1)} = -\frac{1}{t}+\frac{1}{t-1}

\Rightarrow \int \frac{1}{x(x^4+1)}dx =\frac{1}{4}\int \left \{ \frac{-1}{t}+\frac{1}{t-1} \right \}dt

= \frac{1}{4} \left [ -\log|t|+\log|t-1| \right ]+C

= \frac{1}{4}\log\left | \frac{t-1}{t} \right |+C

Back substituting the value of t,

=\frac{1}{4}\log \left | \frac{x^4-1}{x^4} \right | +C

Question:21 Integrate the rational functions \frac{1}{( e ^x-1)} [Hint : Put e ^x= t ]

Answer:

Given function \frac{1}{( e ^x-1)}

So, applying the hint: Putting e^x = t

Then e^x dx= dt

\int \frac{1}{( e ^x-1)}dx = \int\frac{1}{t-1}\times\frac{dt}{t} = \int \frac{1}{t(t-1)}dt


Partial fraction of above equation,

\frac{1}{t(t-1)} = \frac{A}{t}+\frac{B}{(t-1)}

1= A(t-1)+Bt ..............(1)

Now, substituting t = 0\ and\ t = 1 in equation (1), we get

A = -1\ and\ B=1

\Rightarrow \frac{1}{t(t+1)} = -\frac{1}{t}+\frac{1}{t-1}

\implies \int \frac{1}{t(t-1)}dt = \log \left | \frac{t-1}{t} \right |+C

Now, back substituting the value of t,

= \log \left | \frac{e^x-1}{e^x} \right |+C

Question:22 Choose the correct answer \int \frac{x dx }{( x-1)(x-2) } \: \: equals

A ) \log |\frac{(x-1)^2}{x-2}| + C \\\\ B) \log |\frac{(x-2)^2}{x-1}| + C \\\\ C ) \log |(\frac{x-1}{x-2})^2| + C \\\\ D ) \log |{(x-1)^2}({x-2})| + C

Answer:

Given integral \int \frac{x dx }{( x-1)(x-2) }

Partial fraction of above equation,

\frac{x}{(x-1)(x-2)} = \frac{A}{(x-1)}+\frac{B}{(x-2)}

x= A(x+2)+B(x-1) ..............(1)

Now, substituting x = 1\ and\ x = 2 in equation (1), we get

A = -1\ and\ B=2

\therefore \frac{x}{(x-1)(x-2)} = -\frac{1}{(x-1)}+\frac{2}{(x-2)}

\implies \int \frac{x}{(x-1)(x-2)}dx = \int \left \{ \frac{-1}{(x-1)}+\frac{2}{(x-2)} \right \}dx

= -\log|x-1| +2log|x-2| +C

=\log \left | \frac{(x-2)^2}{x-1} \right | +C

Therefore, the correct answer is B.

Question:23 Choose the correct answer \int \frac{dx}{x ( x ^2+1)} \: \: equals

A ) \log |x| - \frac{1}{2} \log ( x^2 +1 ) + C \\\\ B ) \log |x|+ \frac{1}{2} \log ( x^2 +1 ) + C \\\\ C )- \log |x| + \frac{1}{2} \log ( x^2 +1 ) + C \\\\ D ) \frac{1}{2}\log |x| +\log ( x^2 +1 ) + C

Answer:

Given integral \int \frac{dx}{x ( x ^2+1)}

Partial fraction of above equation,

\frac{1}{x ( x ^2+1)} = \frac{A}{x}+\frac{Bx+c}{x^2+1}

1= A(x^2+1)+(Bx+C)x

Now, equating the coefficients of x^2,x, and the constant term, we get

A+B = 0 , C=0 , A=1

We have the values, A = 1\ and\ B=-1,\ and\ C=0

\therefore \frac{1}{x ( x ^2+1)} = \frac{1}{x}+\frac{-x}{x^2+1}

\implies \int \frac{1}{x ( x ^2+1)}dx =\int \left \{ \frac{1}{x}+\frac{-x}{x^2+1}\right \}dx

= \log|x| -\frac{1}{2}\log|x^2+1| +C

Therefore, the correct answer is A.



More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.5

The NCERT Class 12 Maths chapter Integrals covers a total of 12 exercises including one Miscellaneous exercise. Exercise 7.5 Class 12 Maths has a total of 23 main questions along with some few subquestions. In NCERT solutions for Class 12 Maths chapter 7 exercise 7.5 questions difficulty level of questions are of moderate to advanced level which is useful for competitive exams like NEET and JEE Main.

Also Read| Integrals Class 12 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.5

  • The Class 12th Maths chapter 7 exercise is very long. So one should skip some questions to cover maximum syllabus.
  • Practicing exercise 7.5 Class 12 Maths can certainly help students prepare for Board exams and competitive exams.
  • These Class 12 Maths chapter 7 exercise 7.5 solutions can be asked directly in the Board exams.
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Key Features Of NCERT Solutions for Exercise 7.5 Class 12 Maths Chapter 7

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 7.5 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 7.5, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 7.5 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 7.5 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 7.5 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 7.5 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Happy learning!!!

Frequently Asked Question (FAQs)

1. Is there any real life application of Integrals ?

Integration can be used to calculate the centre of gravity, centre of mass etc. which further helps in understanding dynamics of force, pressure etc. in real life. 

2. What is the value of integration of cos x?

Integration of cos x is sin x + c 

3. How much importance does Integrals hold in Board exams ?

In Board exams, questions of around 20 marks are asked directly which can be of great help to students to score well in the examination.  

4. What is the difficulty level of questions asked in Board exams from this exercise ?

This exercise caters to questions of higher difficulty level. Hence it can be said that easy questions are not asked from this exercise. 

5. Mention some topics in Exercise 7.5 Class 12 Maths ?

Topics which are related to finding out integrals of rational functions are included in this exercise. 

6. How many questions are there in this exercise ?

There are 23 main questions in this exercise along with a few subquestions.

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Have a question related to CBSE Class 12th ?
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

You can use them people also used problem

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Divyan 06 August,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

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Yuvan 01 September,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Read Complete Answer
kumardurgesh1802 10 July,2024
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less than 3

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