`int (x-2) / ((x+1)^2 + 4) dx` Find the indefinite integral

Monday, July 13, 2015

We have to evaluate the integral: $\int \frac{x-2}{(x+1)^2+4}dx$

Let $\displaystyle{x}+{1}={u}$

So, $\displaystyle{\left.{d}{x}\right.}={d}{u}$

Hence we have,

$\int \frac{x-2}{(x+1)^2+4}dx=\int \frac{u-3}{u^2+4}du$

$=\int \frac{u}{u^2+2^2}du-\int\frac{3}{u^2+2^2}du$

First we will evaluate $\int \frac{u}{u^2+4}du$

Let $\displaystyle{{u}}^{{2}}+{4}={t}$

So, $\displaystyle{2}{u}{d}{u}={\left.{d}{t}\right.}$

Therefore we can write,

$\int \frac{u}{u^2+4}du=\int \frac{dt}{2t}$

$=\frac{1}{2}ln(t)$

$=\frac{1}{2}ln(u^2+4)$

Now we will evaluate, $\int \frac{3}{u^2+4}du$

$\int \frac{3}{u^2+2^2}du=\frac{3}{2}tan^{-1}(\frac{u}{2})$

Therefore we have,

$\int \frac{x-2}{(x+1)^2+4}dx=\frac{1}{2}ln(u^2+4)-\frac{3}{2}tan^{-1}(\frac{u}{2})+C$

$=\frac{1}{2}ln((x+1)^2+4)-\frac{3}{2}tan^{-1}(\frac{x+1}{2})+C$

$=\frac{1}{2}ln(x^2+2x+5)-\frac{3}{2}tan^{-1}(\frac{x+1}{2})+C$

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