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explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.10 question 4

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Answer: $2 \log |(x-1)|-\frac{1}{x-1}+c$

Hint: Use substitution method to solve this type of integral

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

Solution:

Let $I=\int \frac{2 x-1}{(x-1)^{2}} d x$

Put $x-1=t \Rightarrow d x=d t$ then

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

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

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

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

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

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

$\begin{aligned} &\Rightarrow I=2 \log |t|+\frac{t^{-1}}{-1}+c \\ & \end{aligned}$

$\Rightarrow I=2 \log |(x-1)|-(x-1)^{-1}+c \\$

$\therefore I=2 \log |(x-1)|-\frac{1}{x-1}+c$

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