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