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

$\int \frac{\left(x^2+1\right)}{(x+1)^2} d x=$
  1. $\quad x-2 \log |(x+1)|-\frac{1}{x+1}+c$, where c is a constant of integration.
  2. $\quad x-2 \log |(x+1)|-\frac{2}{x+1}+\mathrm{c}$, where c is a constant of integration.
  3. $\quad x-\log |(x+1)|-\frac{2}{x+1}+c$, where c is a constant of integration.
  4. $\quad x-\log |(x+1)|-\frac{x}{x+1}+c$, where c is a constant of integration.

Solution

Let $\begin{aligned} I & =\int \frac{\left(x^2+1\right)}{(x+1)^2} \mathrm{~d} x \\ & =\int \frac{\left(x^2+2 x+1-2 x\right)}{(x+1)^2} \mathrm{~d} x \\ & =\int \frac{(x+1)^2}{(x+1)^2} \mathrm{~d} x-\int \frac{2 x}{(x+1)^2} \mathrm{~d} x \\ & =\int 1 \mathrm{~d} x-\int \frac{2 x}{(x+1)^2} \mathrm{~d} x \\ & =\int 1 \mathrm{~d} x-\int \frac{2 x+2-2}{(x+1)^2} \mathrm{~d} x \\ & =x-2 \int \frac{(x+1)}{(x+1)^2}+2 \int \frac{1}{(x+1)^2} \mathrm{~d} x \\ & =x-2 \int \frac{1}{x+1} \mathrm{~d} x+2 \int \frac{1}{(x+1)^2} \mathrm{~d} x\end{aligned}$ $=x-2 \log (x+1)-\frac{2}{(x+1)}+c$

Asked in: MHT CET 2024 (04 May Shift 2)

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