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

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

Let `x+1=u`

So, `dx=du`

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 `u^2+4=t`

So, `2udu=dt`

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`