Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 20

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