`int (2x-5) / (x^2+2x + 2) dx` Find or evaluate the integral by completing the square

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`