RD Sharma Class 12 Exercise 18.13 Indefinite Integrals Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 18.13 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.13 Indefinite Integrals Solutions Maths - Download PDF Free Online

Kuldeep MauryaUpdated on 24 Jan 2022, 12:11 PM IST

RD Sharma class 12th exercise 18.13 a students best study guide as an NCERT solution. Practice makes everything perfect and this perfection is required when students will sit for their board exams. Students in class 12 have to finish a huge maths syllabus in a limited time. This might cause them to panic and stress out. However, with the help of RD Sharma class 12 chapter 18 exercise 18.13, they will be able to practice well at home and master the subject quickly. The RD Sharma solutions will soon become their holy grail and help them score high.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.13
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.13

Indefinite Integrals Excercise 18.13 Question 2

Answer: $\frac{1}{8a^{2}}\frac{x^{8}}{\left ( a^{2}-x^{2} \right )^{4}}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{x^{7}}{\left ( a^{2}-x^{2} \right )^{5}} dx$
Solution:
Let $I=\int \frac{x^{7}}{\left ( a^{2}-x^{2} \right )^{5}} dx$
Put $x=a \sin\theta\Rightarrow dx=a\cos\theta d\theta$ …(i) (Differentiating w.r.t to θ)
$\begin{aligned} &\text { Then } \mathrm{I}=\int \frac{a^{7} \sin ^{7} \theta}{\left(a^{2}-a^{2} \sin ^{2} \theta\right)^{5}} \cdot a \cos \theta d \theta \\ &=\int \frac{a^{8} \sin ^{7} \theta \cos \theta}{\left\{a^{2}\left(1-\sin ^{2} \theta\right)\right\}^{5}} d \theta=\int \frac{a^{8} \sin ^{7} \theta \cos \theta}{\left.a^{2 \times 5}\left(\cos ^{2} \theta\right)\right\}^{5}} d \theta \\ &=\int \frac{a^{8} \sin ^{7} \theta \cos \theta}{a^{10} \cos ^{10} \theta} d \theta=\frac{1}{a^{2}} \int \frac{\sin ^{7} \theta}{\cos ^{9} \theta} d \theta \end{aligned}$
$\begin{aligned} &=\frac{1}{a^{2}} \int \frac{\sin ^{7} \theta}{\cos ^{7} \theta} \cdot \frac{1}{\cos ^{2} \theta} d \theta \\ &=\frac{1}{a^{2}} \int \tan ^{7} \theta \cdot \sec ^{2} \theta d \theta \quad \quad\quad\quad\quad\quad\quad\left[\because \frac{\sin \theta}{\cos \theta}=\tan \theta \text { and } \frac{1}{\cos \theta}=\sec \theta\right] \end{aligned}$
Again put$\tan\theta=t\Rightarrow \sec2\theta d\theta=dt$ (on differentiating w.r.t to $\theta$)
$\begin{aligned} &\text { Then } I=\frac{1}{a^{2}} \int t^{7} d t=\frac{1}{a^{2}} \cdot \frac{t^{7+1}}{7+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &I=\frac{1}{a^{2}} \cdot \frac{t^{8}}{8}+c=\frac{1}{8 a^{2}} \cdot t^{8}+c \\ &I=\frac{1}{8 a^{2}} \cdot t^{8}+c \quad\quad\quad\quad\quad[\because t=\tan \theta] \end{aligned}$
From (i)$x=a\sin\theta$
$\Rightarrow \sin \theta=\frac{x}{a}$
and we know that $\sin^{2}\theta+\cos^{2}\theta=1$
$\begin{aligned} &\Rightarrow \cos ^{2} \theta=1-\sin ^{2} \theta \\ &\Rightarrow \cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1-\left(\frac{x}{a}\right)^{2}} \\ &\Rightarrow \cos \theta=\sqrt{1-\frac{x^{2}}{a^{2}}}=\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}=\frac{1}{a} \sqrt{a^{2}-x^{2}} \end{aligned}$
$\begin{aligned} &\therefore \tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\frac{x}{a}}{\frac{1}{a} \sqrt{a^{2}-x^{2}}} \\ &\Rightarrow \tan \theta=\frac{x}{a} \cdot \frac{a}{\sqrt{a^{2}-x^{2}}}=\frac{x}{\sqrt{a^{2}-x^{2}}} \end{aligned}$

$\therefore \tan\theta=\frac{x}{\sqrt{a^{2}-x^{2}}}$ …(iii)

Put the value of tanθ from (iii) in (ii) then
$\begin{aligned} &I=\frac{1}{8 a^{2}}\left(\frac{x}{\sqrt{a^{2}-x^{2}}}\right)^{8}+c \\ &=\frac{1}{8 a^{2}} \frac{x^{8}}{\left(a^{2}-x^{2}\right)^{\frac{8}{2}}}+c \\ &=\frac{1}{8 a^{2}} \frac{x^{8}}{\left(a^{2}-x^{2}\right)^{4}}+c \end{aligned}$

Indefinite Integrals Excercise 18.13 Question 3

Answer: $\frac{x^{2}}{2}+c$
Hint: Use substitution method to solve this integral
Given: $\int \cos \left \{ 2\cot^{-1} \sqrt{\frac{1+x}{1-x}}\right \}dx$
Put$x=\cos2\theta$ …(i)
$\begin{aligned} &\text { Then } I=\int \cos \left\{2 \cot ^{-1} \sqrt{\frac{1+\cos 2 \theta}{1-\cos 2 \theta}}\right\} d x \\ &=\int \cos \left\{2 \cot ^{-1} \sqrt{\frac{2 \cos ^{2} \theta}{2 \sin ^{2} \theta}}\right\} d x \quad\quad\quad\quad\quad\left\{\begin{array}{l} \because 1+\cos 2 A=2 \cos ^{2} A \\ 1-\cos 2 A=2 \sin ^{2} A \end{array}\right\} \\ &=\int \cos \left\{2 \cot ^{-1} \sqrt{\left(\frac{\cos \theta}{\sin \theta}\right)^{2}}\right\} d x \\ &=\int \cos \left\{2 \cot ^{-1} \sqrt{(\cot \theta)^{2}}\right\} d x \quad\quad\quad\quad\quad\quad\left[\because \frac{\cos A}{\sin A}=\cot A\right] \\ &=\int \cos \left\{2 \cot ^{-1} \cdot \cot \theta\right\} d x \end{aligned}$
$\begin{aligned} &=\int \cos \{2 \theta\} d x \quad\left[\because \cot \left(\cot ^{-1} x\right)=x\right] \\ &=\int \cos \left\{\cos ^{-1}(x)\right\} d x \\ &=\int x^{1} d x \quad\left[\begin{array}{c} \because x=\cos 2 \theta \\ 2 \theta=\cos ^{-1}(x) \end{array}\right] \\ &=\frac{x^{1+1}}{1+1}+c \quad\left[\because \cos \left(\cos ^{-1} x\right)=x\right] \\ &=\frac{x^{2}}{2}+c \end{aligned}$

Indefinite Integrals Excercise 18.13 Question 4

Answer: $\frac{-1}{3} \frac{\left(1+x^{2}\right)^{\frac{3}{2}}}{x^{3}}+c \$
Hint: Use substitution method to solve this integral
Given:$\int \frac{\sqrt{1+x^{2}}}{x^{4}}dx$
Solution:
Let $I=\int \frac{\sqrt{1+x^{2}}}{x^{4}}dx$
Putting $x=\tan\theta$ … (i)
$dx=\sec 2\theta d\theta$
Then
$\begin{aligned} &I=\int \frac{\sqrt{1+\tan ^{2} \theta}}{\tan ^{4} \theta} \sec ^{2} \theta d \theta \\ &=\int \frac{\sqrt{\sec ^{2} \theta}}{\tan ^{4} \theta} \cdot \sec ^{2} \theta d \theta \quad\quad\quad\quad\quad\quad\quad\left[\begin{array}{l} \because \sec ^{2} \theta-\tan ^{2} \theta=1 \\ 1+\tan ^{2} \theta=\sec ^{2} \theta \end{array}\right] \\ &=\int \frac{\sec \theta \cdot \sec ^{2} \theta}{\tan ^{4} \theta} d \theta=\int \frac{\sec ^{3} \theta}{\tan ^{4} \theta} d \theta \\ &=\int \frac{1}{\cos ^{3} \theta \cdot \frac{\sin ^{4} \theta}{\cos ^{4} \theta}} d \theta=\int \frac{\cos ^{4} \theta}{\sin ^{4} \theta \cdot \cos ^{3} \theta} d \theta \\ &=\int \frac{\cos \theta}{\sin ^{4} \theta} d \theta \end{aligned}$
Again putting $\sin\theta=t$
$\begin{aligned} &\Rightarrow \cos \theta d \theta=d t \text { then } \\ &I=\int \frac{1}{t^{4}} \cdot d t \\ &=\int t^{-4} d t=\frac{t^{-4+1}}{-4+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=\frac{t^{-3}}{-3}+c \\ &=\frac{-1}{3} t^{-3}+c \\ &=\frac{-1}{3 t^{3}}+c \quad \text { u. (ii) } \end{aligned}$
Also from (i), $x=\tan\theta$
$\begin{aligned} &x^{2}=\tan ^{2} \theta\quad\quad\quad\quad \quad \text { [Squaring both sides] }\\ &\Rightarrow x^{2}=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \quad\quad\quad\quad\quad\quad\quad\quad\left[\tan \theta=\frac{\sin \theta}{\cos \theta}\right]\\ &\Rightarrow x^{2} \cos ^{2} \theta=\sin ^{2} \theta\\ &\Rightarrow x^{2}\left(1-\sin ^{2} \theta\right)=\sin ^{2} \theta \quad\quad\quad\quad\quad\left[\begin{array}{l} \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \Rightarrow \sin ^{2} \theta=1-\cos ^{2} \theta \end{array}\right] \end{aligned}$
$\begin{aligned} &\Rightarrow x^{2}-x^{2} \sin ^{2} \theta=\sin ^{2} \theta\\ &\Rightarrow x^{2}=\sin ^{2} \theta+x^{2} \sin ^{2} \theta=\sin ^{2} \theta\left(1+x^{2}\right)\\ &\Rightarrow \sin ^{2} \theta=\frac{x^{2}}{1+x^{2}}\\ &\Rightarrow \sin \theta=\sqrt{\frac{x^{2}}{1+x^{2}}}\\ &\Rightarrow \sin \theta=\frac{x}{\sqrt{1+x^{2}}} \end{aligned}$
Putting the value of sinθ from (iii) in (ii), we get
$I=\frac{-1}{3}\frac{1}{\left ( \sin\theta \right )^{3}}+c$
$\begin{aligned} &=\frac{-1}{3} \frac{1}{\left(\frac{x}{\sqrt{1+x^{2}}}\right)^{3}}+c \\ &=\frac{-1}{3} \frac{1}{\frac{x^{3}}{\left(\sqrt{1+x^{2}}\right)^{3}}}+c \\ &=-\frac{1}{3} \frac{\left(1+x^{2}\right)^{\frac{3}{2}}}{x^{3}}+c \end{aligned}$

Indefinite Integrals Excercise 18.13 Question 1

Answer:$\frac{x}{\sqrt{a^{2}-x^{2}}}-\sin ^{-1}\left(\frac{x}{a}\right)+c$
Hint: Use substitution method to solve this integral.
Given:$\int \frac{x^{2}}{\left(a^{2}-x^{2}\right)^{\frac{3}{2}}} d x$
Solution:
Let $I=\int \frac{x^{2}}{\left(a^{2}-x^{2}\right)^{\frac{3}{2}}} d x$
Put$x = a \sin u$ …(i)
$dx = a \cos u du$ (on differentiating w.r.t to u)
Then
$\begin{aligned} &I=\int \frac{a^{2} \sin ^{2} u}{\left(a^{2}-a^{2} \sin ^{2} u\right)^{\frac{3}{2}}} \cdot a \cos u d u=\int \frac{a^{3} \sin ^{2} u \cdot \cos u}{\left(a^{2}\left(1-\sin ^{2} u\right)\right\}^{\frac{3}{2}}} d u \\ &=\int \frac{a^{3} \sin ^{2} u \cdot \cos u}{\left(a^{2}\right)^{\frac{3}{2}}\left(1-\sin ^{2} u\right)^{\frac{3}{2}}} d u=\int \frac{a^{3} \sin ^{2} u \cdot \cos u}{\left(a^{3}\right)\left(\cos ^{2} u\right)^{\frac{3}{2}}} d u \\ &=\int \frac{a^{3} \sin ^{2} u \cdot \cos u}{\left(a^{3}\right)\left(\cos ^{2} u\right)^{\frac{3}{2}}} d u \\ &=\int \frac{\sin ^{2} u \cdot \cos u}{(\cos u)^{2 \times \frac{3}{2}}} d u \end{aligned}$
$\begin{aligned} &=\int \frac{\sin ^{2} u \cdot \cos u}{(\cos u)^{3}} d u \\ &=\int \frac{\sin ^{2} u}{(\cos u)^{2}} d u=\int\left(\frac{\sin u}{\cos u}\right)^{2} d u \\ &=\int(\tan u)^{2} d u=\int \tan ^{2} u d u \quad\left[\because \frac{\sin \theta}{\cos \theta}=\tan \theta\right] \end{aligned}$
$\begin{aligned} &=\int\left(\sec ^{2} u-1\right) d u=\int \sec ^{2} u d u-\int 1 d u \\ &{\left[\because 1+\tan ^{2} \theta=\sec ^{2} \theta \Rightarrow \tan ^{2} \theta=\sec ^{2} \theta-1\right]} \\ &I=\tan u-u+c \quad \text { ... (ii) } \quad\quad\quad\quad\left[\because \int 1 d x=x+c\right] \end{aligned}$
From (i) $x = a \sin u$
$\Rightarrow \operatorname{\sin} \mathrm{u}=\frac{x}{a} \Rightarrow u=\sin ^{-1}\left(\frac{x}{a}\right)$ …(iii)
And we know that $1=\sin ^{2} u+\cos ^{2} u$ then
$\begin{aligned} &\Rightarrow \cos ^{2} u=1-\sin ^{2} u \\ &\Rightarrow \cos u=\sqrt{1-\sin ^{2} u}=\sqrt{1-\left(\frac{x}{a}\right)^{2}} \end{aligned}$

Now

$\begin{aligned} &\operatorname{tan} u=\frac{\sin u}{\cos u}=\frac{\frac{x}{a}}{\sqrt{1-\frac{x^{2}}{a^{2}}}}=\frac{\frac{x}{a}}{\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}}=\frac{\frac{x}{a}}{\frac{1}{a} \sqrt{a^{2}-x^{2}}} \\ &\tan u=\frac{x}{a} \cdot \frac{a}{\sqrt{a^{2}-x^{2}}}=\frac{x}{\sqrt{a^{2}-x^{2}}} \\ &\operatorname{tan} u=\frac{x}{\sqrt{a^{2}-x^{2}}} \end{aligned}$ …(iv)
Now put the value of tan u and u from equation (iv) and (iii) in equation (ii), we get
$\mathrm{I}=\frac{x}{\sqrt{a^{2}-x^{2}}}-\sin ^{-1}\left(\frac{x}{a}\right)+c$

Indefinite Integrals Excercise 18.13 Question 5

Answer: $\frac{1}{54}\left[\tan ^{-1}\left(\frac{x+1}{3}\right)+\frac{3(x+1)}{x^{2}+2+10}\right]+c$
Hint: Use substitution method to solve this integral
Given:$\int \frac{1}{\left(x^{2}+2 x+10\right)^{2}} d x \\$
Solution:
$\begin{aligned} &\text { Let } I=\int \frac{1}{\left(x^{2}+2 x+10\right)^{2}} d x \\ &=\int \frac{1}{\left(x^{2}+2 x+1+9\right)^{2}} d x \\ &=\int \frac{1}{\left\{\left(x^{2}+2 x \cdot 1+1\right)+9\right\}^{2}} d x \\ &=\int \frac{1}{\left\{(x+1)^{2}+9\right\}^{2}} d x \end{aligned}$

Putting $x+1=3\tan\theta$ …(i)

$d x=3 \sec ^{2} \theta d \theta$
Then
$\begin{aligned} &I=\int \frac{1}{\left\{(3 \tan \theta)^{2}+9\right\}^{2}} 3 \sec ^{2} \theta d \theta \\ &=\int \frac{3 \sec ^{2} \theta}{\left\{9\left(1+\tan ^{2} \theta\right)\right\}^{2}} d \theta=\int \frac{3 \sec ^{2} \theta}{9^{2}\left(\sec ^{2} \theta\right)^{2}} d \theta \end{aligned}$

$\begin{aligned} &=\int \frac{3 \sec ^{2} \theta}{81 \sec ^{4} \theta} d \theta \quad\left[\begin{array}{l} \because \sec ^{2} \theta-\tan ^{2} \theta=1 \\ \Rightarrow 1+\tan ^{2} \theta=\sec ^{2} \theta \end{array}\right] \\ &=\int \frac{1}{27} \frac{1}{\sec ^{2} \theta} d \theta=\int \frac{1}{27} \cos ^{2} \theta d \theta \quad\left[\because \cos \theta=\frac{1}{\sec \theta}\right] \\ &=\frac{1}{27} \int\left\{\frac{1+\cos 2 \theta}{2}\right\} d \theta \quad\left[\begin{array}{l} \because 2 \cos ^{2} A=1+\cos 2 A \\ \cos ^{2} A=\frac{1+\cos 2 A}{2} \end{array}\right] \end{aligned}$

$\begin{aligned} &=\frac{1}{27}\left\{\int \frac{1}{2} d \theta+\int \frac{1}{2} \cos 2 \theta d \theta\right\} \\ &=\frac{1}{27} \times \frac{1}{2} \int \theta^{0} d \theta+\frac{1}{27} \times \frac{1}{2} \int \cos 2 \theta d \theta \\ &=\frac{1}{54} \cdot \frac{\theta^{0+1}}{0+1}+\frac{1}{54} \frac{\sin 2 \theta}{2}+c\quad\quad\quad\quad[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \text { and } \int \cos a x d x=\frac{\sin a x}{a}+c ]\\ \end{aligned}$
$=\frac{1}{54} \theta+\frac{1}{108} \sin 2 \theta+c$ …(ii)
Also from (i)
$\begin{aligned} &x+1=3 \tan \theta \\ &\tan \theta=\frac{x+1}{3} \Rightarrow \theta=\tan ^{-1}\left(\frac{x+1}{3}\right) \end{aligned}$ … (iii)
And$\tan\theta=\frac{x+1}{3}$
$\begin{array}{ll} \Rightarrow \tan ^{2} \theta=\left(\frac{x+1}{3}\right)^{2} & \text { [Squaring both sides] } \\\\ \Rightarrow \frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\frac{(x+1)^{2}}{9} \quad & {\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]} \\ \\\Rightarrow \sin ^{2} \theta=\frac{(x+1)^{2}}{9} \cdot \cos ^{2} \theta & {\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]} \end{array}$
$\begin{aligned} &\Rightarrow \sin ^{2} \theta=\frac{\left(x^{2}+2 x+1\right) \cos ^{2} \theta}{9} \\ &\Rightarrow \sin ^{2} \theta=\frac{\left(x^{2}+2 x+1\right)}{9}\left(1-\sin ^{2} \theta\right) \quad\left[\begin{array}{l} \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \Rightarrow \cos ^{2} \theta=1-\sin ^{2} \theta \end{array}\right] \\ &\Rightarrow \sin ^{2} \theta=\frac{\left(x^{2}+2 x+1\right)}{9}-\left(\frac{\left(x^{2}+2 x+1\right)}{9}\right) \sin ^{2} \theta \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{x^{2}+2 x+1}{9}=\sin ^{2} \theta+\left(\frac{x^{2}+2 x+1}{9}\right) \sin ^{2} \theta \\ &\Rightarrow \frac{x^{2}+2 x+1}{9}=\sin ^{2} \theta\left\{1+\left(\frac{x^{2}+2 x+1}{9}\right)\right\} \\ &\Rightarrow \frac{x^{2}+2 x+1}{9}=\left(\frac{9+x^{2}+2 x+1}{9}\right) \sin ^{2} \theta=\left(\frac{x^{2}+2 x+10}{9}\right) \sin ^{2} \theta \end{aligned}$
$\begin{aligned} &\Rightarrow \sin ^{2} \theta=\frac{x^{2}+2 x+1}{9} \times \frac{9}{x^{2}+2 x+10}=\frac{(x+1)^{2}}{\left(x^{2}+2 x+10\right)} \quad\quad\quad\quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \\ &\Rightarrow \sin \theta=\sqrt{\frac{(x+1)^{2}}{x^{2}+2 x+10}}=\frac{(x+1)}{\sqrt{x^{2}+2 x+10}} \quad \ldots \text { (iv) } \\ &\quad \text { Since } \sin ^{2} \theta+\cos ^{2} \theta=1 \end{aligned}$

So we can write

$\begin{aligned} &\Rightarrow \cos ^{2} \theta=1-\sin ^{2} \theta \\ &\Rightarrow \cos \theta=\sqrt{1-\sin ^{2} \theta} \\ &\begin{aligned} \Rightarrow \cos \theta &=\sqrt{1-\left[\frac{(x+1)}{\sqrt{x^{2}+2 x+10}}\right]^{2}} \\ &=\sqrt{\frac{x^{2}+2 x+10-\left(x^{2}+2 x+1\right)}{x^{2}+2 x+10}} \end{aligned} \quad\quad\quad\quad\quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \end{aligned}$
$\begin{aligned} &\Rightarrow \cos \theta=\sqrt{\frac{x^{2}+2 x+10-x^{2}-2 x-1}{x^{2}+2 x+10}}\\ &\Rightarrow \cos \theta=\sqrt{\frac{9}{x^{2}+2 x+10}}=\frac{3}{\sqrt{x^{2}+2 x+10}}\quad\quad\quad\quad....(v) \end{aligned}$

Also, we know $\sin 2 \theta=2 \sin \theta \cos \theta$

$\begin{aligned} &\Rightarrow \sin 2 \theta=2 \frac{(x+1)}{\sqrt{x^{2}+2 x+10}} \cdot \frac{3}{\sqrt{x^{2}+2 x+10}} \quad\quad \quad \quad [\text { from (iv) and (v)] }\\ &\Rightarrow \sin 2 \theta=\frac{2.3(x+1)}{x^{2}+2 x+10}=\frac{6(x+1)}{x^{2}+2 x+10}\quad \quad \quad \quad \cdot \cdot \cdot \cdot (vi) \end{aligned}$
Now put the value of θ and sin2θ from equation (iii) and (vi) in (ii) we get
$\begin{aligned} &I=\frac{1}{54} \theta+\frac{1}{108} \sin 2 \theta+c \\ &=\frac{1}{54} \tan ^{-1}\left(\frac{x+1}{3}\right)+\frac{1}{108} \cdot \frac{6(x+1)}{\left(x^{2}+2 x+10\right)}+c \\ &=\frac{1}{54} \tan ^{-1}\left(\frac{x+1}{3}\right)+\frac{1}{54} \frac{3(x+1)}{\left(x^{2}+2 x+10\right)}+c \\ &=\frac{1}{54}\left[\tan ^{-1}\left(\frac{x+1}{3}\right)+\frac{3(x+1)}{\left(x^{2}+2 x+10\right)}\right]+c \end{aligned}$

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