RD Sharma Class 12 Exercise 18.29 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.29 Indefinite Integrals Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.29

Indefinite Integrals Exercise 18.29 Question 1

Answer :

- \frac{1}{3} {(x^{2} - x + 1)}^{\frac{3}{2}} + \frac{3}{8} (2 x - 1 \sqrt{x^{2} - x + 1} + 3 \ln | x + \frac{1}{2} + \sqrt{x^{2} - x + 1} | + C

Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given :

\int (x + 1) \sqrt{x^{2} - x + 1} d x

Solution :

\begin{aligned} x + 1 = a \frac{d}{d x} (x^{2} - x + 1) + b \\ \Rightarrow x + 1 = a (2 x - 1) + b \\ \Rightarrow x + 1 = 2 a x - a + b \end{aligned}

Comparing the coefficient of x and constant terms, we get

\begin{aligned} \Rightarrow 2 a = 1 \Rightarrow a = \frac{1}{2} \\ \Rightarrow b - a = 1 \Rightarrow b = 1 + a \Rightarrow b = 1 + \frac{1}{2} \Rightarrow b = \frac{3}{2} \end{aligned}

\begin{aligned} I = \int (\frac{1}{2} (2 x - 1) + \frac{3}{2}) \sqrt{x^{2} - x + 1} d x \\ I = \int \frac{1}{2} (2 x - 1) \sqrt{x^{2} - x + 1} d x + \int \frac{3}{2} \sqrt{x^{2} - x + 1} d x \end{aligned}

For first integral let

x^{2} - x + 1 = t \Rightarrow (2 x - 1) d x = d t

\begin{aligned} I = \frac{1}{2} \int \sqrt{t} d t + \frac{3}{2} \int \sqrt{x^{2} - x + 1} d x \\ I = \frac{1}{2} \int \sqrt{t} d t + \frac{3}{2} \int \sqrt{x^{2} - 2 x (\frac{1}{2}) + {(\frac{1}{2})}^{2} + 1 - {(\frac{1}{2})}^{2}} d x \end{aligned}

I = \frac{1}{2} \frac{t^{\frac{1}{2} + 1}}{\frac{3}{2}} + \frac{3}{2} \int \sqrt{{(x - \frac{1}{2})}^{2} + \frac{3}{4}} d x

I = \frac{1}{3} t^{3 / 2} + \frac{3}{2} \int \sqrt{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x

Usinf formula,

\int \sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log | x + \sqrt{x^{2} + a^{2}} | + C

I = \frac{1}{3} t^{3 / 2} + \frac{3}{2} (\frac{x - \frac{1}{2}}{2} (\sqrt{({(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2})}) + \frac{{(\frac{\sqrt{3}}{2})}^{2}}{2} \log | x - \frac{1}{2} + \sqrt{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} |) + C

\begin{aligned} I = \frac{1}{3} t^{3 / 2} + \frac{3}{2} (\frac{2 x - 1}{4} \sqrt{x^{2} - x + 1} + \frac{3}{8} \log | x - \frac{1}{2} + \sqrt{x^{2} - x + 1} |) + C \\ I = \frac{1}{3} t^{3 / 2} + \frac{3}{2} (\frac{2 x - 1}{4} \sqrt{x^{2} - x + 1} + \frac{3}{8} \log | x - \frac{1}{2} + \sqrt{x^{2} - x + 1} |) + C \end{aligned}

I = \frac{1}{3} {(x^{2} - x + 1)}^{3 / 2} + \frac{3}{8} (2 x - 1) \sqrt{x^{2} - x + 1} + \frac{9}{16} \log [x - \frac{1}{2} + \sqrt{x^{2} - x + 1}] + c

Indefinite Integrals Exercise 18.29 Question 2

Answer :

\frac{1}{6} {(2 x^{2} + 3)}^{\frac{3}{2}} + \frac{x}{2} \sqrt{x^{2} - x + 1} + \frac{3 \sqrt{2}}{4} \log | \frac{\sqrt{2 x} + \sqrt{2 x^{2} + 3}}{\sqrt{3}} | + C

Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given :

\int (x + 1) \sqrt{2 x^{2} + 3} d x

Solution :

Let I = \int x \sqrt{2 x^{2} + 3} d x + \int \sqrt{2 x^{2} + 3} d x

\begin{aligned} I_{1} = \int x \sqrt{2 x^{2} + 3} d x \\ Let 2 x^{2} + 3 = t \\ 4 x d x = d t \\ x \cdot d x = \frac{d t}{4} \end{aligned}

\begin{aligned} I_{1} = \int \sqrt{t} \frac{d t}{4} \\ = \frac{1}{4} \int t^{\frac{1}{2}} d x \end{aligned}

= \frac{1}{4} \frac{t^{1 / 2 + 1}}{1 / 2 + 1} [\int x^{n} d x = \frac{x^{n} + 1}{n + 1}]

\begin{aligned} = \frac{1}{4} \times 2 \frac{t^{\frac{3}{2}}}{3} \\ = \frac{t^{\frac{3}{2}}}{6} = \frac{{(2 x^{2} + 3)}^{\frac{3}{2}}}{6} + C_{1} \end{aligned}

....(i)

I_{2} = \int \sqrt{2 x^{2} + 3} d x

Use the formula :

\int \sqrt{x^{2} + a^{2}} d x = [\frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log | x + \sqrt{x^{2} + a^{2}} | + C]

\begin{aligned} I_{2} = \int \sqrt{2 (x^{2} + \frac{3}{2})} d x \\ I_{2} = \sqrt{2} \int \sqrt{x^{2} + \frac{3}{2}} d x \\ I_{2} = \sqrt{2} \int \sqrt{x^{2} + {(\frac{\sqrt{3}}{\sqrt{2}})}^{2}} d x \end{aligned}

I_{2} = \sqrt{2} [\frac{x}{2} \sqrt{x^{2} + \frac{3}{2}} + \frac{3}{2 \times 2} \log | x + \sqrt{x^{2} + \frac{3}{2}} |] + C_{2}

....(ii)
Adding (i) and (ii) ;

I = I_{1} + I_{2}

I = \int x \sqrt{2 x^{2} + 3} d x + \int \sqrt{2 x^{2} + 3} d x

\begin{aligned} = \frac{{(2 x^{2} + 3)}^{\frac{3}{2}}}{6} + \frac{x}{\sqrt{2}} \sqrt{x^{2} + \frac{3}{2}} + \frac{3}{2 \sqrt{2}} \log | x + \sqrt{x^{2} + \frac{3}{2}} | + C \\ = \frac{1}{6} {(2 x^{2} + 3)}^{\frac{3}{2}} + \frac{x}{\sqrt{2}} \frac{\sqrt{2 x^{2} + 3}}{\sqrt{2}} + \frac{3}{2 \sqrt{2}} \log | x + \frac{\sqrt{2 x^{2} + 3}}{\sqrt{2}} | + C \end{aligned}

\begin{aligned} = \frac{1}{6} {(2 x^{2} + 3)}^{\frac{3}{2}} + \frac{x}{2} \sqrt{2 x^{2} + 3} + \frac{3}{2 \sqrt{2}} \frac{\sqrt{2}}{\sqrt{2}} \log | \frac{\sqrt{2} x + \sqrt{2 x^{2} + 3}}{\sqrt{2}} | + C \\ \int (x + 1) \sqrt{2 x^{2} + 3} d x = \frac{1}{6} {(2 x^{2} + 3)}^{\frac{3}{2}} + \frac{x}{2} \sqrt{2 x^{2} + 3} + \frac{3 \sqrt{2}}{4} \log | \frac{\sqrt{2} x + \sqrt{2 x^{2} + 3}}{\sqrt{2}} | + c \end{aligned}

Indefinite Integrals Exercise 18.29 Question 3

Answer :

I = - \frac{2}{3} {(2 + 3 x - x^{2})}^{3} - (\frac{2 x - 3}{2}) \sqrt{2 + 3 x - x^{2}} - \frac{17}{4} \sin^{- 1} (\frac{2 x - 3}{\sqrt{17}}) + C

Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given :

\int (2 x - 5) \sqrt{2 + 3 x - x^{2}} d x

Solution :

\begin{aligned} 2 x - 5 = - (5 - 2 x) = - (2 + 3 - 2 x) = - 2 - (3 - 2 x) \\ I = \int (- 2 - (3 - 2 x)) \sqrt{2 + 3 x - x^{2}} d x \\ I = \int - 2 \sqrt{2 + 3 x - x^{2}} d x - \int (3 - 2 x) \sqrt{2 + 3 x - x^{2}} d x \end{aligned}

Again,

2 + 3 x - x^{2}

\begin{aligned} = - (x^{2} - 3 x - 2) \\ = - (x^{2} - 2 \cdot x \cdot \frac{3}{2} + \frac{9}{4} - 2 - \frac{9}{4}) \\ = - ({(x - \frac{3}{2})}^{2} - \frac{17}{4}) \\ = ({(\sqrt{\frac{17}{4}})}^{2} - {(x - \frac{3}{2})}^{2}) \end{aligned}

Now,

I = - 2 \int [\sqrt{{(\sqrt{\frac{17}{4}})}^{2} - {(x - \frac{3}{2})}^{2}}] dx - \int (3 - 2 x) \sqrt{2 + 3 x - x^{2}} d x

For the second integral :
Let,

2 + 3 x - x^{2} = t^{2}

\Rightarrow [(3 - 2 x) d x = 2 t d t]

Use the formula :

\int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a} + C

\begin{aligned} I = - 2 [\frac{(x - \frac{3}{2})}{2} \sqrt{\frac{17}{4} - {(x - \frac{3}{2})}^{2}} + \frac{\frac{17}{4}}{2} \sin^{- 1} \frac{(x - \frac{3}{2})}{\frac{17}{4}}] - \int 2 t^{2} d t \\ I = - (\frac{2 x - 3}{2}) \sqrt{2 + 3 x - x^{2}} - \frac{17}{4} \sin^{- 1} (\frac{2 x - 3}{\sqrt{17}}) - \frac{2}{3} {(2 + 3 x - x^{2})}^{3} + C \\ I = - \frac{2}{3} {(2 + 3 x - x^{2})}^{3} - (\frac{2 x - 3}{2}) \sqrt{2 + 3 x - x^{2}} - \frac{17}{4} \sin^{- 1} (\frac{2 x - 3}{\sqrt{17}}) + C \end{aligned}

Indefinite Integrals Exercise 18.29 Question 4

Answer :

\frac{1}{3} {(x^{2} + x + 1)}^{\frac{3}{2}} + \frac{3 (2 x + 1)}{8} \sqrt{x^{2} + x + 1} + \frac{9}{16} \log | (x + \frac{1}{2}) + \sqrt{x^{2} + x + 1} | + C

Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given :

\int (x + 2) \sqrt{x^{2} + x + 1} d x

Solution :

\begin{aligned} Let (x + 2) = a \frac{d}{d x} (x^{2} + x + 1) + b \\ \Rightarrow x + 2 = a (2 x + 1) + b \\ \Rightarrow x + 2 = 2 a x + a + b \end{aligned}

Now comparing the coefficients of x and the constant term, we get

\begin{aligned} 2 a = 1 \Rightarrow a = \frac{1}{2} and \\ a + b = 2 \Rightarrow b = 2 - a = 2 - \frac{1}{2} = \frac{3}{2} \\ I = \int (\frac{1}{2} (2 x + 1) + \frac{3}{2}) \sqrt{x^{2} + x + 1} d x \\ I = \int \frac{1}{2} (2 x + 1) \sqrt{x^{2} + x + 1} d x + \int \frac{3}{2} \sqrt{x^{2} + x + 1} d x \end{aligned}

For the first integral, let

x^{2} + x + 1 = t \Rightarrow (2 x + 1) d x = d t

\begin{aligned} I = \frac{1}{2} \int \sqrt{t} d t + \frac{3}{2} \int \sqrt{x^{2} + 2 (x) (\frac{1}{2}) + {(\frac{1}{2})}^{2} + 1 - \frac{1}{4}} d x \\ I = \frac{1}{2} \frac{t^{\frac{1}{2} + 1}}{\frac{3}{2}} + \frac{3}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} d x [\int x^{n} d x = \frac{x^{n} + 1}{n + 1} + C] \end{aligned}

I = \frac{t^{\frac{3}{2}}}{3} + \frac{3}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x

(Use the formula: \sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log | x + \sqrt{x^{2} + a^{2}} | + C)

\begin{aligned} I = \frac{{(x^{2} + x + 1)}^{\frac{3}{2}}}{3} + \frac{3}{2} [\frac{(x + \frac{1}{2})}{2} \sqrt{x^{2} + x + 1} + \frac{{(\frac{\sqrt{3}}{2})}^{2}}{2} \log | (x + \frac{1}{2}) + \sqrt{x^{2} + x + 1} |] + C \\ I = \frac{1}{3} {(x^{2} + x + 1)}^{3 / 2} + \frac{3 (2 x + 1)}{8} \sqrt{x^{2} + x + 1} + \frac{9}{16} \log | (x + \frac{1}{2}) + \sqrt{x^{2} + x + 1} | + C \end{aligned}

Indefinite Integrals Exercise 18.29 Question 5

Answer:

\frac{4}{3} {(x^{2} + x + 1)}^{\frac{3}{2}} + \frac{3 (2 x + 1)}{8} \sqrt{x^{2} + x + 1} + \frac{27}{8} \log | (x + \frac{1}{2}) + \sqrt{x^{2} + x + 1} | + C

Hint: To solve the given integration, we express the linear team as a derivative of quadratic into constant plus another constant
Given:

I = \int (4 x + 1) \sqrt{x^{2} - x - 2} d x

Solution: Let,

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

\begin{aligned} \Rightarrow 4 x + 1 = M (2 x - 1) + N \\ \Rightarrow 4 x + 1 = 2 M x - M + N \end{aligned}

Now comparing the coefficients of x and the constant term, we get

\begin{aligned} 2 M = 4 \Rightarrow M = 2 and - M + N = 1 \Rightarrow - 2 + N = 1 \Rightarrow N = 3 \\ I = 2 \int (2 x - 1) \sqrt{x^{2} - x - 2} d x + 3 \int \sqrt{x^{2} - x - 2} d x \end{aligned}

For first integral, let,

x^{2} - x - 2 = t \Rightarrow (2 x - 1) d x = d t

For second integral,

x^{2} - x - 2 = x^{2} - 2 \cdot x \cdot \frac{1}{2} + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} - 2 = {(x - \frac{1}{2})}^{2} - \frac{9}{4} = {(x - \frac{1}{2})}^{2} - {(\frac{3}{2})}^{2}

So, the integral becomes

I = 2 \int \sqrt{t} d t + 3 \int \sqrt{{(x - \frac{1}{2})}^{2} - {(\frac{3}{2})}^{2}} d x

Use the formula :

[\int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \log | x + \sqrt{x^{2} - a^{2}} |] + C

\begin{aligned} I = 2 \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + 3 [\frac{(x - \frac{1}{2})}{2} \sqrt{x^{2} - x - 2} - \frac{\frac{9}{4}}{2} \log | (x - \frac{1}{2}) + \sqrt{{(x - \frac{1}{2})}^{2} - {(\frac{3}{2})}^{2}} |] + C \\ I = 2 \times 2 \frac{t^{\frac{3}{2}}}{3} + \frac{3}{2} [(x - \frac{1}{2}) \sqrt{x^{2} - x - 2} - \frac{9}{4} \log | (x - \frac{1}{2}) + \sqrt{x^{2} - x - 2} |] + C \end{aligned}

I = \frac{4}{3} {(x^{2} - x - 2)}^{3 / 2} + \frac{3}{4} (2 x - 1) \sqrt{x^{2} x - 2} - \frac{27}{8} \log | (x - \frac{1}{2}) + \sqrt{x^{2} - x - 2} | + C

Indefinite Integrals Exercise 18.29 Question 6

Answer :

I = \frac{1}{6} {(2 x^{2} - 6 x + 5)}^{3 / 2} - \frac{1}{2} [\frac{2 x - 3}{2} \sqrt{x^{2} - 3 x + \frac{5}{2}} + \frac{11}{8} \log | \frac{2 x - 3}{\sqrt{2}} + \sqrt{2 x^{2} - 6 x + 5} |] + C

Hint: We solve this integration by qualitative derivation.
Given:

\int (x - 2) \sqrt{2 x^{2} - 6 x + 5} d x

Let,

x - 2 = a \frac{d}{d x} (2 x^{2} - 6 x + 5) + b

\begin{aligned} \Rightarrow x - 2 = a (4 x - 6) + b \\ \Rightarrow x - 2 = 4 a x + b - 6 a \end{aligned}

Now comparing the coefficients of x and the constant term, we get

\begin{aligned} 4 a = 1 \Rightarrow a = \frac{1}{4} and \\ b - 6 a = - 2 \Rightarrow b = 6 (\frac{1}{4}) - 2 \Rightarrow b = (- \frac{1}{2}) \\ I = \int [\frac{1}{4} (4 x - 6) + (- \frac{1}{2})] \sqrt{2 x^{2} - 6 x + 5} d x \end{aligned}

I = \int \frac{1}{4} (4 x - 6) \sqrt{2 x^{2} - 6 x + 5} d x + \int - \frac{1}{2} \sqrt{2 x^{2} - 6 x + 5} d x

For the first integral : Let,

2 x^{2} - 6 x + 5 = t \Rightarrow (4 x - 6) d x = d t

I = \frac{1}{4} \int \sqrt{t} d t + (- \frac{1}{2}) \int \sqrt{(\sqrt{2} x)^{2} - 2 (\sqrt{2} x) (\frac{3}{\sqrt{2}}) + {(\frac{3}{\sqrt{2}})}^{2} + 5 - \frac{9}{4}} d x

\begin{aligned} I = \frac{1}{4} \frac{t^{\frac{1}{2} + 1}}{3 / 2} - \frac{1}{2} \int \sqrt{{(\sqrt{2} x - \frac{3}{\sqrt{2}})}^{2} + \frac{11}{4}} d x [\int x^{n} d x = \frac{x^{n} + 1}{n + 1} + C] \\ I = \frac{1}{2 \times 3} t^{3 / 2} - \frac{1}{2} \int \sqrt{{(\sqrt{2} x - \frac{3}{\sqrt{2}})}^{2} + {(\frac{\sqrt{11}}{2})}^{2}} d x \end{aligned}

Use the formula :

[\sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log | x + \sqrt{x^{2} + a^{2}} |]

\begin{aligned} I = \frac{1}{6} {(2 x^{2} - 6 x + 5)}^{\frac{3}{2}} - \frac{1}{2} [\frac{(\sqrt{2} x - \frac{3}{\sqrt{2}})}{2} \sqrt{2 x^{2} - 6 x + 5} + \frac{\frac{11}{4}}{2} \log | (\sqrt{2} x - \frac{3}{\sqrt{2}}) + \sqrt{2 x^{2} - 6 x + 5} |] + C \\ I = \frac{1}{6} {(2 x^{2} - 6 x + 5)}^{3 / 2} - \frac{1}{2} [\frac{2 x - 3}{2} \sqrt{x^{2} - 3 x + \frac{5}{2}} + \frac{11}{8} \log | \frac{2 x - 3}{\sqrt{2}} + \sqrt{2 x^{2} - 6 x + 5} |] + C \end{aligned}

Indefinite Integrals Exercise 18.29 Question 7

Answer :

I = \frac{(x^{2} + x + 1)}{2} + \frac{1}{2} [\frac{2 x + 1}{4} \sqrt{x^{2} + x + 1} + \frac{3}{8} \log | (x + \frac{1}{2}) + \sqrt{x^{2} + x + 1} |] + C

Hint : To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given :

\int (x + 1) \sqrt{x^{2} + x + 1} d x

Solution :

\begin{aligned} I = \frac{1}{2} \int (2 x + 2) \sqrt{x^{2} + x + 1} d x \\ I = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x + 1} d x + \frac{1}{2} \int 1 \sqrt{x^{2} + x + 1} d x \end{aligned}

For thr first integral: Let,

x^{2} + x + 1 = t^{2} \Rightarrow (2 x + 1) d x = 2 t d t

\begin{aligned} I = \frac{1}{2} \int 2 t d t + \frac{1}{2} \int \sqrt{x^{2} + 2 (x) (\frac{1}{2}) + {(\frac{1}{2})}^{2} + 1 - \frac{1}{4}} d x \\ I = \int t d t + \frac{1}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} d x \end{aligned}

I = \frac{t^{1 + 1}}{1 + 1} + \frac{1}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x [\int x^{n} d x = \frac{x^{n} + 1}{n + 1} + C]

\begin{aligned} I = \frac{t^{2}}{2} + \frac{1}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x \\ (Use the formula: \sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log | x + \sqrt{x^{2} + a^{2}} | + C) \end{aligned}

\begin{aligned} I = \frac{(x^{2} + x + 1)}{2} + \frac{1}{2} [\frac{(x + \frac{1}{2})}{2} \sqrt{x^{2} + x + 1} + \frac{{(\frac{\sqrt{3}}{2})}^{2}}{2} \log | (x + \frac{1}{2}) + \sqrt{x^{2} + x + 1} |] + C \\ I = \frac{(x^{2} + x + 1)}{2} + \frac{1}{2} [\frac{2 x + 1}{4} \sqrt{x^{2} + x + 1} + \frac{3}{8} \log | (x + \frac{1}{2}) + \sqrt{x^{2} + x + 1} |] + C \end{aligned}

Indefinite Integrals Exercise 18.29 Question 8

Answer :

I = \frac{2}{3} {(x^{2} + 4 x + 3)}^{3 / 2} - \frac{(x + 2)}{2} \sqrt{x^{2} + 4 x + 3} + \frac{1}{2} \log | (x + 2) + \sqrt{x^{2} + 4 x + 3} | + C

Hint:To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given :

\int (2 x + 3) \sqrt{x^{2} + 4 x + 3} d x

Solution :
Let,

x^{2} + 4 x + 3 = u^{2}

\begin{aligned} \Rightarrow (2 x + 4) d x = 2 u d u \\ I = \int (2 x + 3) \sqrt{x^{2} + 4 x + 3} d x \\ I = \int (2 x + 4 - 1) \sqrt{x^{2} + 4 x + 3} d x \end{aligned}

\begin{aligned} I = \int (2 x + 4) \sqrt{x^{2} + 4 x + 3} d x - \int \sqrt{x^{2} + 4 x + 3} dx \\ I = \int u (2 u) du - \int \sqrt{(x + 2)^{2} - 1^{2}} d x [x^{2} + 4 x + 3 = (x + 2)^{2} - 1] \end{aligned}

Use the formula :

[\int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \log | x + \sqrt{x^{2} - a^{2}} |] + C

And

[\int x^{n} d x = \frac{x^{n} + 1}{n + 1} + C]

\begin{aligned} I = 2 \frac{u^{3}}{3} - [\frac{x + 2}{2} \sqrt{(x + 2)^{2} - 1} - \frac{1}{2} \log | (x + 2) + \sqrt{{(x^{2} + 2)}^{2} - 1} |] + C \\ I = \frac{2}{3} {(x^{2} + 4 x + 3)}^{3 / 2} - \frac{(x + 2)}{2} \sqrt{x^{2} + 4 x + 3} + \frac{1}{2} \log | (x + 2) + \sqrt{x^{2} + 4 x + 3} | + C \end{aligned}

Indefinite Integrals Exercise 18.29 Question 9

Answer :

I = \frac{2}{3} {(x^{2} - 4 x + 3)}^{3 / 2} - \frac{(x - 2)}{2} \sqrt{x^{2} - 4 x + 3} - \frac{1}{2} \log | (x - 2) + \sqrt{x^{2} - 4 x + 3} | + C

Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given:

\int (2 x - 5) \sqrt{x^{2} - 4 x + 3} d x

\begin{aligned} I = \int [[(2 x - 4) - 1] \sqrt{x^{2} - 4 x + 3}] d x \\ I = \int (2 x - 4) \sqrt{x^{2} - 4 x + 3} d x - \int \sqrt{x^{2} - 4 x + 3} d x \end{aligned}

For the first integral :
Let

x^{2} - 4 x + 3 = t

\begin{aligned} \Rightarrow 2 x - 4 = \frac{d t}{d x} \\ \Rightarrow (2 x - 4) d x = d t \end{aligned}

\begin{aligned} I = \int \sqrt{t} d t - \int \sqrt{x^{2} - 4 x + 3} d x \\ I = \int t^{\frac{1}{2}} d t - \int \sqrt{(x - 2)^{2} - (1)^{2}} d x [x^{2} - 4 x + 3 = (x - 2)^{2} - 1] \end{aligned}

Use the formula :

[\int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \log | x + \sqrt{x^{2} - a^{2}} |] + C

And

[\int x^{n} d x = \frac{x^{n} + 1}{n + 1} + C]

\begin{aligned} I = \frac{t^{1 / 2} + 1}{\frac{1}{2} + 1} - [\frac{x - 2}{2} \sqrt{(x - 2)^{2} - 1} - \frac{1}{2} \log | (x - 2) + \sqrt{(x - 2)^{2} - 1} |] + C \\ I = \frac{2}{3} {(x^{2} - 4 x + 3)}^{3 / 2} - \frac{(x - 2)}{2} \sqrt{x^{2} - 4 x + 3} - \frac{1}{2} \log | (x - 2) + \sqrt{x^{2} - 4 x + 3} | + C \end{aligned}

Indefinite Integrals Exercise 18.29 Question 10

Answer :

I = \frac{1}{3} {(x^{2} + x)}^{3 / 2} - \frac{1}{8} (2 x + 1) \sqrt{x^{2} + x} + \frac{1}{16} \log | (x + \frac{1}{2}) + \sqrt{x^{2} + x} | + C

Hint : To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given :

\int x \sqrt{x^{2} + x} d x

Let,

x = a \frac{d}{d x} (x^{2} + x) + b

\begin{aligned} \Rightarrow x = a (2 x + 1) + b \\ \Rightarrow x = 2 x a + a + b \end{aligned}

Comparing the coefficient of x and the constant terms, we get

\begin{aligned} \Rightarrow 2 a = 1 and a + b = 0 \\ \Rightarrow a = 1 / 2 \Rightarrow b = - a \\ \Rightarrow b = - \frac{1}{2} \\ I = \int (\frac{1}{2} (2 x + 1) + (- \frac{1}{2})) \sqrt{x^{2} + x} d x \end{aligned}

I = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x} d x + \int (- \frac{1}{2}) \sqrt{x^{2} + x} d x

For the first integral : Let

x^{2} + x = t \Rightarrow (2 x + 1) d x = d t

I = \frac{1}{2} \int \sqrt{t} d t + (- \frac{1}{2}) \int \sqrt{x^{2} + 2 (x) (\frac{1}{2}) + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} d x

Use the formula :

[\int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \log | x + \sqrt{x^{2} - a^{2}} |] + C

And

[\int x^{n} d x = \frac{x^{n} + 1}{n + 1} + C]

\begin{aligned} I = \frac{1}{2} \frac{\frac{1}{2} + 1}{\frac{1}{2} + 1} - \frac{1}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} d x \\ I = \frac{1}{3} t^{\frac{3}{2}} - \frac{1}{2} [\frac{(x + \frac{1}{2})}{2} \sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} - \frac{\frac{1}{4}}{2} \log | (x + \frac{1}{2}) + \sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} |] + C \end{aligned}

\begin{aligned} I = \frac{1}{3} {(x^{2} + x)}^{3 / 2} - \frac{1}{2} [\frac{(2 x + 1)}{4} \sqrt{x^{2} + x} - \frac{1}{8} \log | (x + \frac{1}{2}) + \sqrt{x^{2} + x} |] + C \\ I = \frac{1}{3} {(x^{2} + x)}^{3 / 2} - \frac{1}{8} (2 x + 1) \sqrt{x^{2} + x} + \frac{1}{16} \log | (x + \frac{1}{2}) + \sqrt{x^{2} + x} | + C \end{aligned}

Indefinite Integrals Exercise 18.29 Question 11

Answer :

I = \frac{1}{3} (x + 3 x - 18)^{3 / 2} - \frac{9}{8} (2 x + 3) \sqrt{x^{2} + 3 x - 18} + \frac{729}{16} \log | (x + \frac{3}{2}) + \sqrt{x^{2} + 3 x - 18} | + C

Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given :

\int (x - 3) \sqrt{x^{2} + 3 x - 18} d x

Solution :

x - 3 = A + B \frac{d}{d x} (x^{2} + 3 x - 18)

\Rightarrow x - 3 = A + B (2 x + 3)

Comparing the coefficient of x and the constant terms, we get

\Rightarrow 1 = 2 B and \Rightarrow - 3 = A + 3 B

\Rightarrow B = 1 / 2 \Rightarrow - 3 - 3 B = A

\begin{aligned} \Rightarrow - 3 - \frac{3}{2} = A \\ \Rightarrow A = - \frac{9}{2} \end{aligned}

\begin{aligned} I = \int [- \frac{9}{2} + \frac{1}{2} (2 x + 3)] \sqrt{x^{2} + 3 x - 18} d x \\ I = - \frac{9}{2} \int \sqrt{x^{2} + 3 x - 18} d x + \frac{1}{2} \int (2 x + 3) \sqrt{x^{2} + 3 x - 18} d x \end{aligned}

For the second integral:
Let

x^{2} + 3 x - 18 = t

\Rightarrow (2 x + 3) d x = d t

Use the formula :

[\int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \log | x + \sqrt{x^{2} - a^{2}} |] + C

And

[\int x^{n} d x = \frac{x^{n} + 1}{n + 1} + C]

I = - \frac{9}{2} [\frac{(x + \frac{3}{2})}{2} \sqrt{x^{2} + 3 x - 18} - \frac{81}{4 \times 2} \log | (x + \frac{3}{2}) + \sqrt{x^{2} + 3 x - 18} |] + \frac{1}{2} \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + C

\begin{aligned} I = - \frac{9}{2} [(\frac{2 x + 3}{4}) \sqrt{x^{2} + 3 x - 18} - \frac{81}{8} \log | (x + \frac{3}{2}) + \sqrt{x^{2} + 3 x - 18} | + \frac{1}{3} t^{\frac{3}{2}}] + C \\ I = \frac{1}{3} (x + 3 x - 18)^{3 / 2} - \frac{9}{8} (2 x + 3) \sqrt{x^{2} + 3 x - 18} + \frac{729}{16} \log | (x + \frac{3}{2}) + \sqrt{x^{2} + 3 x - 18} | + C \end{aligned}

Indefinite Integrals Exercise 18.29 Question 12

Answer :

I = - \frac{1}{3} {(3 - 4 x - x^{2})}^{\frac{3}{2}} + \frac{1}{2} (x + 2) \sqrt{3 - 4 x - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + C

Hint: To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant
Given :

\int (x + 3) \sqrt{3 - 4 x - x^{2}} d x

Solution :

\begin{aligned} I = \int (x + 2 + 1) \sqrt{3 - 4 x - x^{2}} d x \\ I = \int (x + 2) \sqrt{3 - 4 x - x^{2}} d x + \int \sqrt{3 - 4 x - x^{2}} d x \\ Let, 3 - 4 x - x^{2} = u^{2} \end{aligned}

\begin{aligned} \Rightarrow (- 4 - 2 x) d x = 2 u d u \\ \Rightarrow - 2 (x + 2) d x = 2 u d u \\ \Rightarrow (x + 2) d x = - u d u \end{aligned}

\begin{aligned} I = - \int u \sqrt{u^{2}} d u + \int \sqrt{- (x^{2} + 4 x + 4 - 4) + 3} d x \\ I = - \int u^{2} d u + \int \sqrt{- (x + 2)^{2} + 7} d x \end{aligned}

Use the formula :

\int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a} + C

And

[\int x^{n} d x = \frac{x^{n} + 1}{n + 1} + C]

\begin{aligned} I = - \frac{u^{3}}{3} + \int \sqrt{(\sqrt{7})^{2} - (x + 2)^{2}} d x \\ I = \frac{- {(3 - 4 x - x^{2})}^{3 / 2}}{3} + \frac{x + 2}{2} \sqrt{3 - 4 x - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + C \end{aligned}

I = - \frac{1}{3} {(3 - 4 x - x^{2})}^{\frac{3}{2}} + \frac{1}{2} (x + 2) \sqrt{3 - 4 x - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + C

Indefinite Integrals Exercise 18.29 Question 13

Answer :

I = - \frac{1}{2} {(4 - 3 x - 2 x^{2})}^{\frac{3}{2}} - \frac{5}{4} (\frac{2 x + 3}{2 \sqrt{2}} \sqrt{4 - 3 x - 2 x^{2}} + \frac{17}{4} \sin^{- 1} \frac{(2 x + 3)}{\sqrt{17}}) + C

Hint: To solve the given integration, we express the linear team as a derivative of quadratic into constant plus another constant
Given :

\int (3 x + 1) \sqrt{4 - 3 x - 2 x^{2}} d x

Solution :

\begin{aligned} Let, 3 x + 1 = a \frac{d}{d x} (4 - 3 x - 2 x^{2}) + b \\ \Rightarrow 3 x + 1 = a (- 3 - 4 x) + b \\ \Rightarrow 3 x + 1 = - 4 a x + b - 3 a \end{aligned}

Comparing the coefficient of x and the constant terms, we get

4 a = - 3 \Rightarrow a = - \frac{3}{4}

and

\begin{aligned} b - 3 a = 1 \Rightarrow b = 1 + 3 (- \frac{3}{4}) \Rightarrow b = - \frac{5}{4} \\ I = \int [- \frac{3}{4} (- 3 - 4 x) - \frac{5}{4}] \sqrt{4 - 3 x - 2 x^{2}} d x \end{aligned}

I = \int (- \frac{3}{4}) (- 3 - 4 x) \sqrt{4 - 3 x - 2 x^{2}} d x + \int - \frac{5}{4} \sqrt{4 - 3 x - 2 x^{2}} d x

For the first integral :

\begin{aligned} Let 4 - 3 x - 2 x^{2} = t \\ \Rightarrow (- 3 - 4 x) d x = d t \end{aligned}

I = - \frac{3}{4} \int \sqrt{t} d t - \frac{5}{4} \int \sqrt{4 + {(\frac{3}{\sqrt{2}})}^{2} - [(\sqrt{2} x)^{2} + 2 \sqrt{2} x (\frac{3}{\sqrt{2}}) + {(\frac{3}{\sqrt{2}})}^{2}]} d x

Use the formula :

\int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a} + C

And

[\int x^{n} d x = \frac{x^{n} + 1}{n + 1} + C]

\begin{aligned} I = - \frac{3}{4} \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \frac{5}{4} \int \sqrt{\frac{17}{2} - {(\sqrt{2} x + \frac{3}{\sqrt{2}})}^{2}} d x \\ I = - \frac{3}{4} \frac{t^{\frac{3}{2}}}{\frac{3}{2}} - \frac{5}{4} \sqrt{{(\frac{\sqrt{17}}{\sqrt{2}})}^{2} - {(\sqrt{2} x + \frac{3}{\sqrt{2}})}^{2}} d x \end{aligned}

\begin{aligned} I = - \frac{1}{2} {(4 - 3 x - 2 x^{2})}^{\frac{3}{2}} - \frac{5}{4} (\frac{\sqrt{2} x + \frac{3}{\sqrt{2}}}{2} \sqrt{4 - 3 x - 2 x^{2}} + \frac{17}{2} \frac{1}{2} \sin^{- 1} \frac{(\sqrt{2} x + \frac{3}{\sqrt{2}})}{\frac{\sqrt{17}}{\sqrt{2}}}) + C \\ I = - \frac{1}{2} {(4 - 3 x - 2 x^{2})}^{\frac{3}{2}} - \frac{5}{4} (\frac{2 x + 3}{2 \sqrt{2}} \sqrt{4 - 3 x - 2 x^{2}} + \frac{17}{4} \sin^{- 1} \frac{(2 x + 3)}{\sqrt{17}}) + C \end{aligned}

Indefinite Integrals Exercise 18.29 Question 14

Answer :

\frac{- 2}{9} {(10 - 4 x - 3 x^{2})}^{\frac{3}{2}} + \frac{11}{8} (3 x + 2) \sqrt{10 - 4 x + 3 x^{2}} + \frac{187}{9 \sqrt{3}} \sin^{- 1} (\frac{3 x + 2}{\sqrt{34}}) + C

Hint: To solve the given integration, we express the linear team as a derivative of quadratic into constant plus another constant
Given:

\int (2 x + 5) \sqrt{10 - 4 x - 3 x^{2}} d x

Solution :

\int (2 x + 5) \sqrt{10 - 4 x - 3 x^{2}} d x

\begin{aligned} Let, (2 x + 5) = A \frac{d}{d x} (10 - 4 x - 3 x^{2}) + B \\ (2 x + 5) = A (- 4 - 6 x) + B \\ (2 x + 5) = - 4 A + B - 6 A x \end{aligned}

Comparing the coefficient of x and the constant terms, we get

\begin{aligned} 6 A = - 2 \Rightarrow A = - \frac{1}{3} and \\ - 4 A + B = 5 \Rightarrow B = 5 + 4 A = 5 + 4 (- \frac{1}{3}) = \frac{11}{3} \\ I = - \frac{1}{3} \int (- 4 - 6 x) \sqrt{10 - 4 x - 3 x^{2}} d x \\ I = - \frac{1}{3} \int (- 6 x - 4) \sqrt{10 - 4 x - 3 x^{2}} d x \end{aligned}

\begin{aligned} I = - \frac{1}{3} [\int (- 6 x - 4) \sqrt{10 - 4 x - 3 x^{2}} d x + \int \frac{11}{3} \sqrt{10 - 4 x - 3 x^{2}} d x] \\ I = - \frac{1}{3} \int (- 6 x - 4) \sqrt{10 - 4 x - 3 x^{2}} d x + (\frac{11}{3}) \int \sqrt{10 - 4 x - 3 x^{2}} d x \end{aligned}

For the first integral:

\begin{aligned} Let, 10 - 4 x - 3 x^{2} = t \\ \Rightarrow (- 6 x - 4) d x = d t \\ I = - \frac{1}{3} \int \sqrt{t} d t + \frac{11}{3} \times \sqrt{3} \int \sqrt{\frac{10}{3} - \frac{4}{3} x - x^{2}} d x \end{aligned}

Use the formula :

\int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a} + C

And

[\int x^{n} d x = \frac{x^{n} + 1}{n + 1} + C]

\begin{aligned} I = - \frac{1}{3} \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + \frac{11}{3} \sqrt{3} \int \sqrt{(\frac{10}{3}) - {x^{2} + 2 \frac{2}{3} x + {(\frac{2}{3})}^{2}} + {(\frac{2}{3})}^{2}} d x \\ I = - \frac{2}{9} t^{3 / 2} + \frac{11}{\sqrt{3}} \int \sqrt{{(\frac{\sqrt{34}}{3})}^{2} - {(x + \frac{2}{3})}^{2}} d x \end{aligned}

\begin{aligned} I = - \frac{2}{9} {(10 - 4 x - 3 x^{2})}^{\frac{3}{2}} + \frac{11}{\sqrt{3}} [\frac{1}{2} (x + \frac{2}{3}) \sqrt{{(\frac{\sqrt{34}}{3})}^{2} - {(x + \frac{2}{3})}^{2}} + \frac{34 / 9}{2} \sin^{- 1} [\frac{x + \frac{2}{3}}{\frac{\sqrt{34}}{3}}]] + C \\ I = - \frac{2}{9} {(10 - 4 x - 3 x^{2})}^{\frac{3}{2}} + \frac{11}{\sqrt{3}} [\frac{(x + \frac{2}{3}) {(\frac{10}{3} - \frac{4 x}{3} - x^{2})}^{\frac{1}{2}}}{2}] + \frac{187}{9 \sqrt{3}} \sin^{- 1} [\frac{3 x + 2}{\sqrt{34}}] + C \end{aligned}

I = - \frac{2}{9} {(10 - 4 x - 3 x^{2})}^{\frac{3}{2}} + \frac{11}{18} (3 x + 2) \sqrt{10 - 4 x - 3 x^{2}} + \frac{187}{9 \sqrt{3}} \sin^{- 1} [\frac{3 x + 2}{\sqrt{34}}] + C

Rd Sharma class 12 chapter 18 exercise 18.29 has around 14 inquiries, including its subparts, and it consolidates themes like: -

Assessment of integrals by utilizing mathematical replacements like first need to solve the quadratic equation then proceed with the differentiation and integration
Questions based on the Formula of Integration

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