Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 36

We have to evaluate the integral: \int \frac{2x-5}{x^2+2x+2}dx
We can write the integral as:
\int \frac{2x-5}{x^2+2x+2}dx=\int\frac{2x-5}{(x+1)^2+1}dx
Let x+1=t
So, dx=dt
Now we can write the integral as:
\int \frac{2x-5}{(x+1)^2+1}dx=\int \frac{2(t-1)-5}{t^2+1}dt
=\int \frac{2t-7}{t^2+1}dt
=\int \frac{2t}{t^2+1}dt-\int\frac{7}{t^2+1}dt --------------->(1)

Now we will first evaluate the integral \int \frac{2t}{t^2+1}dt

Let t^2+1=u
So, 2tdt=du
Hence we can write,
\int \frac{2tdt}{t^2+1}=\int \frac{du}{u}
=ln(u)
=ln(t^2+1)

Now we will evaluate the second integral : \int \frac{7}{t^2+1}dt
\int \frac{7}{t^2+1}dt=7\int \frac{1}{t^2+1}dt
=7tan^{-1}(t)

Substituting both these integral results in (1) we get,
\int \frac{2x-5}{x^2+2x+2}dx=ln(t^2+1)-7tan^{-1}(t)+C where C is a constant
=ln((x+1)^2+1)-7tan^{-1}(x+1)+C
=ln(x^2+2x+2)-7tan^{-1}(x+1)+C