#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions  Question 35 Maths Textbook Solution.

$a=\frac{-1}{10}, b=\frac{2}{5}$

Given:

$\int \frac{1}{(x+2)\left(x^{2}+1\right)} d x=a \log \left|1+x^{2}\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+C$

Hint:

Using partial fraction and $\int \frac{1}{x} d x$and $\int \frac{1}{1+x^{2}} d x$

Explanation:

Let $\mathrm{I}=\int \frac{1}{(x+2)\left(x^{2}+1\right)} d x$

Let $\frac{1}{(x+2)\left(x^{2}+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+1}$                        $\left [ Eq.(i)) \right ]$

Multiplying by $(x+2)\left(x^{2}+1\right)$, we get

$I=A\left(x^{2}+1\right)+(B x+C)(x+2)$

Putting x = -2

\begin{aligned} &1=A(5)+(-2 B+C)(-2+2) \\ &\Rightarrow 1=5 A+0 \\ &\Rightarrow A=\frac{1}{5} \end{aligned}

Putting x = 0

\begin{aligned} &1=A(1)+(B(0)+C)(0+2) \\ &\Rightarrow 1=\frac{1}{5}+2 C \\ &\Rightarrow C=\frac{2}{5} \end{aligned}

Putting x = +1

\begin{aligned} &1=A(1+1)+(B+C)(1+2) \\ &\Rightarrow 1=2\left(\frac{1}{5}\right)+\left(B+\frac{2}{5}\right)(3) \\ &\Rightarrow 1=\frac{2}{5}+3 B+\frac{6}{5} \\ &\Rightarrow 3 B=1-\frac{8}{5} \\ &\Rightarrow 3 B=\frac{-3}{5} \\ &\Rightarrow B=\frac{-1}{5} \end{aligned}

$\therefore \text { From }[E q \cdot(i)]$

\begin{aligned} &\frac{1}{(x+2)\left(x^{2}+1\right)}=\frac{\frac{1}{5}}{x+2}+\frac{\left(\frac{-1}{5}\right) x+\left(\frac{2}{5}\right)}{x^{2}+1}\\ &\therefore \int \frac{1}{(x+2)\left(x^{2}+1\right)} d x=\frac{1}{5} \int \frac{1}{x+2} d x+\left(\frac{-1}{5}\right) \cdot \frac{1}{2} \int \frac{2 x}{x^{2}+1} d x+\frac{2}{5} \int \frac{1}{x^{2}+1} d x\\ &\therefore I=\frac{1}{5} \log |x+2|-\frac{1}{10} \log \left|1+x^{2}\right|+\frac{2}{5} \tan ^{-1} x+C \end{aligned}$\left [ Eq.(ii)) \right ]$

Acc. to given

$I=a \log \left|1+x^{2}\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|$                                                                              $\left [ Eq.(iii)) \right ]$

\begin{aligned} &\therefore \text { From } E q \cdot(i i) \text { and } E q \cdot(\text { iii }) \text { we get }\\ &a=\frac{-1}{10}, b=\frac{2}{5} \end{aligned}