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

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

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

$A = -\frac{1}{6},\ \text{and}\ B = \frac{1}{6}$

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

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

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

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

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

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

$A = 1,\ B = -5,\ \text{and}\ C = 4$

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

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

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

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

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

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

$A =2,\ B=3$

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

Now, 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 function $\frac{x }{( x^2+1 )( x-1)}$

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

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

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

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

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

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

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

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

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

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

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

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

[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

$A = 1\ \text{and}\ B = -1$

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

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

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

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

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

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

$A = 0,\ B = -2,\ C = 0,\ \text{and}\ D = 6$

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

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

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

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

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

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

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

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

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 $t = 0\ \text{and}\ t = 1$ in equation (1), we get

$A = -1\ \text{and}\ B = 1$

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

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

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

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

Back substituting the value of t,

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

Question 21: Integrate the rational function $\frac{1}{( e ^x-1)}$ [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 $t = 0\ \text{and}\ t = 1$ in equation (1), we get

$A = -1\ \text{and}\ B = 1$

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

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

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

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

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

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

$D)\ \log \left| (x - 1)^2 (x - 2) \right| + C$

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 $x = 1\ \text{and}\ x = 2$ in equation (1), we get

$A = -1\ \text{and}\ B = 2$

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

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

$= -\text{log}|x - 1| + 2\text{log}|x - 2| + C$

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

Therefore, the correct answer is $\log \left| \frac{(x - 2)^2}{x - 1} \right| + C$.

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, $A = 1\ \text{and}\ B = -1,\ \text{and}\ C = 0$

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

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

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

Therefore, the correct answer is $\log |x| - \frac{1}{2} \log (x^2 + 1) + C$.