#### Explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.10 question 6

Answer:  $(x+1)+\log |(x+1)|+\frac{1}{x+1}+c$

Hint: Use substitution method to solve this type of integral

Given:  $\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x$

Solution: let  $I=\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x$

Substitute $x+1=t \Rightarrow d x=d t$  then

$\Rightarrow I=\int \frac{(t-1)^{2}+3(t-1)+1}{t^{2}} d t \qquad(\because x=t-2)$

\begin{aligned} &\Rightarrow I=\int \frac{t^{2}-2 t+1+3 t-3+1}{t^{2}} d t \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \\ & \end{aligned}

$\Rightarrow I=\int \frac{t^{2}+t-1}{t^{2}} d t \\$

$\Rightarrow I=\int \frac{t^{2}}{t^{2}}+\frac{t}{t^{2}}-\frac{1}{t^{2}} d t$

\begin{aligned} &\Rightarrow I=\int\left(1+\frac{1}{t}-t^{-2}\right) d t \\ & \end{aligned}

$\Rightarrow I=\int t^{0} d t+\int \frac{1}{t} d t-\int t^{-2} d t$

$\Rightarrow I=\frac{t^{0+1}}{0+1}+\log |t|-\frac{t^{-2+1}}{-2+1}+c \qquad\left[\int \frac{1}{t} d t=\log |t|+c \& \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$

\begin{aligned} &\Rightarrow I=t+\log |t|-\frac{t^{-1}}{-1}+c \\ & \end{aligned}

$\Rightarrow I=t+\log |t|+\frac{1}{t}+c \\$

$\therefore I=(x+1)+\log |(x+1)|+\frac{1}{x+1}+c$