RD Sharma Class 12 Exercise 18.31 Indefinite Integrals Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 18.31 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 11:03 AM IST

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## Indefinite Integrals Excercise:18.31

Indefinite Integrals Exercise18.31 Question 1

Answer: The required value of the integral is,
$I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{3} x}\right)+c$
Hint: Use the formula $\int \frac{1}{x^{2}+a^{2}}dx=\frac{1}{a}\tan ^{-1}\frac{x}{a}+c$
Given:
$\int \frac{\left ( x^{2}+1 \right )}{\left ( x^{4}+x^{2}+1 \right )}dx$
Solution: The equation can be written as
$I=\int \frac{1+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x$ [ dividing $x^{2}$ both denominator and denominator]
$=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x^{2}}\right)^{2}+3} d x$ [Making the perfect square as $(a+b)^{2}$ ]
Let $x-\frac{1}{x}=t$
Now differentiating both side w.r.t t
\begin{aligned} \left(1+\frac{1}{x^{2}}\right) &=d t \\ I &=\int \frac{1}{t^{2}+3} d t \end{aligned}
On using standard identity we get, $I=\frac{1}{\sqrt{3}}\left ( \frac{t}{\sqrt{3}} \right )+c$
Substituting $t=x-\frac{1}{x}$ we get,
\begin{aligned} &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\left(x-\frac{1}{x}\right)}{\sqrt{3}}\right)+c \\ &I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{3} x}\right)+c \end{aligned}
So, the required value of the integration is,
$I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{3} x}\right)+c$

Indefinite Integrals Exercise18.31 Question 2

Answer: The required value of the integral is,
$I=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\cot \theta-1}{\sqrt{2} \cot \theta}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{\cot \theta+1-\sqrt{2} \cot \theta}{\cot \theta+1+\sqrt{2} \cot \theta}\right|+c$

Hint: Use the identity formula $\int \frac{1}{x^{2}-a^{2}}dx=\frac{1}{2a}\log\left ( \frac{x-a}{x+a} \right )+c$

Given:$I=\int \sqrt{\cot \theta d \theta}$
Solution: Suppose $\cot \theta \; asx^{2}$
Differentiating the above equation with respect to θ.
\begin{aligned} -\operatorname{cosec}^{2} \theta d \theta &=2 x d x \\ d \theta &=-\frac{2 x}{1+\cot ^{2} \theta} d \theta \\ d \theta &=-\frac{2 x}{1+x^{4}} d x \\ I &=\int-\frac{2 x}{1+x^{4}} d x \quad\quad\quad\quad\quad \ldots .1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta \\ \end{aligned}
Re-writing the given equation as,
$I=\int \frac{1+\frac{1}{x^{2}}+1-\frac{1}{x^{2}}}{\frac{1}{x^{2}}+x^{2}} d x-\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+2} d x-\int \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-2}$ [Making the perfect square as $(a+b)^{2}$ ]
\begin{aligned} &\text { Let, } x-\frac{1}{x}=t \text { and } x+\frac{1}{x}=z \text { so, } \\ &I=-\int \frac{d t}{t^{2}+2}-\int \frac{d z}{z^{2}-2} \end{aligned}

On using identity,

$=-\frac{1}{2}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+c$
The required value of the integral is,
$=-\frac{1}{2}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+c$
Now, substitute$x=\sqrt{}\cot \theta$into the above equation we get,
$I=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\cot \theta-1}{\sqrt{2} \cot \theta}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{\cot \theta+1-\sqrt{2} \cot \theta}{\cot \theta+1+\sqrt{2} \cot \theta}\right|+c$
Where c is the integrating constant.

Indefinite Integrals Exercise18.31 Question 3

Answer: The required value of the integral is,
$I=\frac{1}{3 \sqrt{2}}\left(\frac{x-\frac{9}{x}}{3 \sqrt{2}}\right)+c$
Hint: Use the formula$\int \frac{1}{x^{2}+a^{2}}dx=\frac{1}{a}\tan^{-1}\frac{x}{a}+c$
Given:$\int \frac{\left(x^{2}+9\right)}{\left(x^{4}+81\right)} d x$
Solution: The equation can be written as
$I=\int \frac{1+\frac{9}{x^{2}}}{x^{2}+\frac{81}{x^{2}}} d x$
$=\int \frac{1+\frac{9}{x^{2}}}{\left(x-\frac{9}{x}\right)^{2}+18} d x$
Let, $x-\frac{9}{x}=t$ then,
Differentiating the above equation both sides.
$\left(1+\frac{9}{x^{2}}\right) d x=d t$ then the obtained equation become
$\int \frac{d t}{t^{2}+(3 \sqrt{2})^{2}}$
Now, using the identity rule,
\begin{aligned} \int \frac{1}{x^{2}+1} d x &=\tan ^{-1}(x)+c \\ &=\frac{1}{3 \sqrt{2}} \tan ^{-1}\left(\frac{t}{3 \sqrt{2}}\right)+c \end{aligned}
Now, re-substituting $t=x-\frac{9}{x}$ then,
\begin{aligned} I &=\frac{1}{3 \sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{9}{x}}{3 \sqrt{2}}\right)+c \\ &=\frac{1}{3 \sqrt{2}} \tan ^{-1}\left(\frac{x^{2}-9}{3 \sqrt{2} x}\right)+c \end{aligned}
Where c is the integrating constant.

Indefinite Integrals Exercise18.31 Question 4

Answer: The required value of the integral is,
$I=\frac{1}{2}\left[\frac{1}{\sqrt{3}}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)-\frac{1}{2} \log \left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|\right]$
Hint: Use the identity formula
$\int \frac{1}{x^{2}+1} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c \text { and } \int \frac{d x}{x^{2}-1}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c$
Given:
$\int \frac{1}{x^{4}+x^{2}+1} d x$
Solution: The given equation can be written as,
$I=\int \frac{\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x$ [dividing $x^{2}$ in both denminator and numerator]
\begin{aligned} &=\frac{1}{2} \int \frac{1+\frac{1}{x^{2}}+\frac{1}{x^{2}}-1}{x^{2}+1+\frac{1}{x^{2}}} d x \\ &=\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x+\int \frac{-1+\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-1} d x\right] \\ &=\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+3} d x+\int \frac{-1+\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-1} d x\right] \end{aligned} [Making the perfect square as$\left ( a+b \right )^{2}$ ]
Re-writing the given equation as,
\begin{aligned} &\text { Let, } x-\frac{1}{x}=t \text { and } x+\frac{1}{x}=z \text { so } \\ &\begin{aligned} \left(1+\frac{1}{x^{2}}\right) d x &=d t \text { and }\left(1-\frac{1}{x^{2}}\right) d x=d z \\ I &=\frac{1}{2}\left[\int \frac{d t}{t^{2}+3}-\int \frac{d z}{z^{2}-1}\right] \end{aligned} \end{aligned}
On using identity,
\begin{aligned} \int \frac{1}{x^{2}+1} d x &=\tan ^{-1} x \text { and } \int \frac{d z}{z^{2}-1}=\frac{1}{2} \log \left|\frac{z-1}{z+1}\right|+c \\ I &=\frac{1}{2}\left[\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{3}}\right)-\frac{1}{2} \log \left|\frac{z-1}{z+1}\right|\right] \end{aligned}
Now, re-substituting $t=x-\frac{1}{x}and\: z=x+\frac{1}{x}$
The required value of the integral is,
\begin{aligned} I &=\frac{1}{2}\left[\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)-\frac{1}{2} \log \left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|\right]+c \\ &=\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{x^{2}-1}{x \sqrt{3}}\right)-\frac{1}{4} \log \left|\frac{x^{2}-x+1}{x^{2}+x+1}\right|+c \end{aligned}
Where c is the integrating constant.

Indefinite Integrals Exercise18.31 Question 5

Answer: The required value of the integral is,
$I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)-\sqrt{3} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+c$
Hint: Use the identity formula$\int \frac{1}{x^{2}+1} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$
Given:$I=\int \frac{x^{2}-3 x+1}{x^{4}+x^{2}+1} d x$
Solution: The given equation can be written as,
\begin{aligned} I &=\int \frac{1-\frac{3}{x}+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x \\ &=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+3} d x-\int \frac{3 x}{x^{4}+x^{2}+1} d x \end{aligned} [dividing $x^{2}$ i both numerator and denominator]
[ Making the perfect square as $\left ( a+b \right )^{2}$ ]
Now, substituting$x-\frac{1}{x}=t\: and \: x^{2}=z$ then,
\begin{aligned} &\left(1+\frac{1}{x^{2}}\right) d x=d \tan d 2 x d x=d z \\ &I=\int \frac{d t}{t^{2}+3}-\frac{3}{2} \int \frac{d z}{z^{2}+z+1} \end{aligned}
We knows
$\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$
\begin{aligned} &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{3}}\right)+C_{1}-\frac{3}{2} \int \frac{d z}{\left(z+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t}{\sqrt{3}}\right)+C_{1}-\frac{3}{2} \times \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\left|z+\frac{1}{2}\right|}{\frac{\sqrt{3}}{2}}\right)+C_{2} \end{aligned}
\begin{aligned} &t=x-\frac{1}{4} \text { and } z=x^{2} \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)+C_{1}-\sqrt{3} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+C_{2} \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{3} x}\right)-\sqrt{3} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+C \end{aligned}

Indefinite Integrals Exercise18.31 Question 6

Answer: The required value of the integration is,
$\tan ^{-1}\left(\frac{x^{2}-1}{x}\right)+c$
Hint: Use the identity formula $\int\frac{1}{x^{2}+a^{2}}dx=\frac{1}{a}\tan ^{-1}\frac{x}{a}+c$
Given: The given integral is $I=\int \frac{x^{2}+1}{x^{4}-x^{2}+1}dx$
Solution:
The given equation can be written as,
\begin{aligned} I &=\int \frac{x^{2}+1}{\frac{x^{2}}{\left(\frac{x^{4}-x^{2}+1}{x^{2}}\right)}} d x \\ &=\int \frac{\left(1+\frac{1}{x^{2}}\right) d x}{\left(x^{2}-1+\frac{1}{x^{2}}\right)} \\ &=\int \frac{\left(1+\frac{1}{x^{2}}\right) d x}{x^{2}+\frac{1}{x^{2}}+1-2} \end{aligned} [divinding $x^{2}$ on both numerator and denominator ]
$=\int \frac{\left(1+\frac{1}{x^{2}}\right) d x}{\left(x-\frac{1}{x}\right)^{2}+1^{2}}$ [Making the perfect square of $\left ( a+b \right )^{2}$ ]
Now, substitute $x-\frac{1}{x}=t$ then,
$\left (1+\frac{1}{x^{2}} \right )dx=dt$
Now, solving the obtained equation,
\begin{aligned} I &=\int \frac{d t}{t^{2}+1} \\ &=\tan ^{-1} t+c \end{aligned}
Re-substitute $t=x-\frac{1}{x}$ into the above equation.
\begin{aligned} I &=\tan ^{-1}\left(x-\frac{1}{x}\right)+c \\ &=\tan ^{-1}\left(\frac{x^{2}-1}{x}\right) \end{aligned}
Thus the required value of the given integration is,
$\tan ^{-1}\left(\frac{x^{2}-1}{x}\right)+c$

Indefinite Integrals Exercise18.31 Question 7

Answer: The required value of the integration is,
$\frac{1}{2 \sqrt{2}} \log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+c$
Hint: Use the identity formula $\int \frac{1}{x^{2}+a^{2}}dx=\frac{1}{a}\tan^{-1}\frac{x}{a}+c$
Given:
$I=\int \frac{x^{2}-1}{x^{4}+1} d x$

Solution: Given equation can be written as
$I=\frac{x^{2}-1}{x^{4}+1} d x$
Divide numerator and denominator by $x^{2}$
\begin{aligned} I &=\int \frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}} d x \\ &=\int \frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+2-2} d x \\ &=\int \frac{\left(1-\frac{1}{x^{2}}\right)}{\left(x+\frac{1}{x}\right)^{2}-2} d x \end{aligned}
Let$x+\frac{1}{x}=t$
$\left ( 1-\frac{1}{x^{2}}\right )dx=dt$ differentiating on both sides with respect to t,
\begin{aligned} I &=\int \frac{d t}{t^{2}-2} \\ &=\int \frac{d t}{(t)^{2}-\left(\sqrt{2)^{2}}\right.} \\ &=\frac{1}{2 \sqrt{2}} \log \left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|+c \end{aligned}
Re- substitute$t=x+\frac{1}{x}$ into the above equation.
$\frac{1}{2 \sqrt{2}} \log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+c$
Where c is an integrating constant.

Indefinite Integrals Exercise18.31 Question 8

Answer: The required value of the integral is,
$\frac{1}{3} \tan ^{-1}\left(\frac{x^{2}-1}{3 x}\right)+c$
Given:
$\int \frac{x^{2}+1}{x^{4}+7 x^{2}+1} d x$
Hint: The given integral function can be converted into
$\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$
Solution: Rewrite the given equation as.
$\int \frac{x^{2}+1}{x^{4}+7 x^{2}+1} d x$
Divide numerator and denominator by $x^{2}$
\begin{aligned} I &=\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+7} d x \\ &=\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+7+2-2} dx \\ &=\int \frac{1+\frac{1}{x^{2}}}{\left(x^{2}+\frac{1}{x^{2}}-2\right)+9} d x \\ &=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+9} d x \end{aligned}
$let(x-\frac{1}{x})=t\Rightarrow (1+\frac{1}{x^{2}})dx=dt$
Evaluating the obtained integration.
\begin{aligned} &I=\int \frac{d t}{t^{2}+3^{2}} \\ &I=\frac{1}{3} \tan -1\left(\frac{t}{3}\right)+c \\ &I=\frac{1}{3} \tan ^{-1}\left(\frac{\left.x-\frac{1}{x}\right)+c}{3}\right)+c \\ &I=\frac{1}{3} \tan ^{-1}\left(\frac{x^{2}-1}{3 x}\right)+c &\text { (Substituting } t \text { as }\left(x-\frac{1}{x}\right) \text { ) } \\ \end{aligned}

Indefinite Integrals Exercise18.31 Question 9

$\frac{1}{\sqrt{3}}\left(\frac{x^{2}-1}{\sqrt{3} x}\right)-\frac{2}{\sqrt{3}}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+c$
Given:
$\int \frac{x^{2}-1}{x^{4}+x^{2}+1}$
Hint:
The given integral function can be converted into
$\int \frac{dx}{x^{2}+a^{2}}=\frac{1}a{}\tan^{-1}\frac{x}{a}+c$
Solution
\begin{aligned} I &=\int \frac{(x-1)^{2}}{x^{4}+x^{2}+1} d x \\ &=\int \frac{x^{2}-2 x+1}{x^{4}+x^{2}+1} d x \\ &=\int \frac{x^{2}+1}{x^{4}+x^{2}+1} d x+\int \frac{-2 x}{x^{4}+x^{2}+1} d x \end{aligned}
\begin{aligned} \text { let } x^{2} &=t \\ 2 x d x &=d t \\ I &=\int \frac{x^{2}\left(1+\frac{1}{x^{2}}\right)}{x^{2}\left(x^{2}+1+\frac{1}{x^{2}}\right)} d x+\int \frac{-d t}{t^{2}+t+1} \text { (substituting } t \text { as }x^{2}) \\ &=\int \frac{\left(1+\frac{1}{x^{2}}\right)}{\left(x-\frac{1}{x^{2}}\right)^{2}+2+1} d x-\int \frac{1}{\left(t+\frac{1}{2}\right)^{2}-\frac{1}{4}+1} d t \end{aligned}
\begin{aligned} &\text { let } x-\frac{1}{x}=p \\ &\left(1+\frac{1}{x^{2}}\right) d x=d p \\ &I=\int \frac{d p}{p^{2}+3}-\int \frac{d t}{\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}}\left(\text { substituting } p \text { as } x-\frac{1}{x}\right) \\ &=\int \frac{1}{p^{2}+(\sqrt{3})^{2}} d p-\int \frac{1}{\left(t+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d t \end{aligned}
Using the identity
$\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \frac{x}{a}+c$
\begin{aligned} I &=\frac{1}{\sqrt{3}} \tan ^{-1} \frac{p}{\sqrt{3}}-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) \\ &\left.=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right) \text { (substituting } p \text { as }\left(x-\frac{1}{x}\right) \text { and } t \text { as } x^{2}\right) \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{3} x}\right)-\frac{2}{3} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+c \end{aligned}
So, the required value of the given integral is,
$\frac{1}{\sqrt{3}}\left(\frac{x^{2}-1}{\sqrt{3} x}\right)-\frac{2}{\sqrt{3}}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+c$
where c is integrating constant.

Indefinite Integrals Exercise18.31 Question 10

$\frac{1}{2 \sqrt{5}}\left(\frac{x^{2}-1}{\sqrt{5} x}\right)-\frac{1}{2}\left(\frac{x^{2}+1}{x}\right)+c$
Given:
$\int \frac{1}{x^{4}+3 x^{2}+1} d x$
Hint:
The given integral function can be converted into
$\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$
Solution
\begin{aligned} I &=\int \frac{1}{x^{4}+3 x^{2}+1} d x \\ &=\int \frac{\frac{1}{x^{2}}}{x^{2}+3+\frac{1}{x^{2}}} d x \end{aligned}(Numerator and denominator divided by $x^{2}$)
\begin{aligned} &=\frac{1}{2} \int \frac{\left(1+\frac{1}{x^{2}}\right)-\left(1-\frac{1}{x^{2}}\right)}{x^{2}+3+\frac{1}{x^{2}}} d x \\ &=\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+5} d x-\int \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}+1} d x\right] \end{aligned}
\begin{aligned} \text { Assume } t &=x-\frac{1}{x} \text { and } z=x+\frac{1}{x} \\ d t &=\left(1+\frac{1}{x^{2}}\right) d x \text { and } d z=\left(1-\frac{1}{x^{2}}\right) d x \\ &=\frac{1}{2}\left[\int \frac{d t}{t^{2}+5}-\int \frac{d z}{z^{2}+1}\right] \end{aligned}
Using identity $\int \frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\frac{x}{a}+c$
$I=\frac{1}{2\sqrt{5}}\tan-\left ( \frac{t}{\sqrt{5}} \right )-\frac{1}{2}\tan^{-1}z$
Substituting $t \: as\: x-\frac{1}{x}and\: z\: as\: x+\frac{1}{x}$
\begin{aligned} &I=\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{5}}\right)-\frac{1}{2} \tan ^{-1}\left(x+\frac{1}{x}\right)+c \\ &\frac{1}{2 \sqrt{5}}\left(\frac{x^{2}-1}{\sqrt{5} x}\right)-\frac{1}{2}\left(\frac{x^{2}+1}{x}\right)+c \end{aligned}
Where c is integrating constant.

Indefinite Integrals Exercise18.31 Question 11

$I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)+c$
Hint:The given integral function can be converted into,
$\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$
Given:
$I=\int \frac{1}{\sin ^{4} x+\sin ^{2} x \cos ^{2} x+\cos ^{4} x}$
Solution
\begin{aligned} I &=\int \frac{1}{\sin ^{4} x+\sin ^{2} x \cos ^{2} x+\cos ^{4} x} \\ &=\int \frac{\left(\tan ^{2 } x+1\right) \sec ^{2} x}{\tan ^{4} x+\tan ^{2} x+1} d x \end{aligned} (dividing N and D by $\cos ^{4}x$)
$= \int \frac{t^{2}+1}{t^{4}+t^{2}+1}dt$ where $\tan x= t$
$=\int \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}+1} d t$ (Dividing N and D by $t^{2}$)
putting\; t-\frac{1}{t}=z\; so\; that \left(1+\frac{1}{t^{2}}\right) d t=d \; z\; and \; t^{2}+\frac{1}{t^{2}}=z^{2}+2\\\\ \begin{aligned} I &=\int \frac{d z}{z^{2}+(\sqrt{3})^{2}} \\ &=\frac{1}{\sqrt{3}} \tan ^{-} \frac{z}{\sqrt{3}}+c \\ &=\frac{1}{\sqrt{3}} \tan ^{-}\left(\frac{t^{2}-1}{\sqrt{3} t}\right)+c \end{aligned}
The required value of the given integration is,
$I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)+c$
where c is an integrating constant.

Indefinite Integrals Exercise18.32 Question 12

Answer : - $\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C$
Hint :- Use substitution method to solve this integral.
Given :- $\int \frac{1}{(x-1) \sqrt{x+2}} d x$
Sol : - Let $I=\int \frac{1}{(x-1) \sqrt{x+2}} d x$
Put $x+2=t^{2}\Rightarrow dx=2tdt$
\begin{aligned} &I=\int \frac{1}{\left(t^{2}-2-1\right) \sqrt{t^{2}}} \quad 2 t d t \quad\left(\because x=t^{2}-2\right) \\ &=2 \int \frac{1}{\left(t^{2}-3\right) t} t d t \\ &=2 \int \frac{1}{t^{2}-3} d t \end{aligned}
$\begin{array}{ll} =2 \cdot \frac{1}{2 \sqrt{3}} \log \left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|+C & \left(\because \int \frac{1}{x^{2}-a^{2}} d x=1 / 2 a \log \left|\frac{x-a}{x+a}\right|+C\right) \\\\ =\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C \quad & (\because t=\sqrt{x+2}) \text { Ans.. } \end{array}$

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RD Sharma Chapter-wise Solutions

1. What number of exercises are there in RD Sharma Class 12 chapter 18?

There are around 32 exercises in RD Sharma Class 12 Chapter 18.

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