0

I need to solve $$\int \frac{2x+1}{x^2-2x+2}dx$$

I have noticed that the Numerator is almost the derivate of the Denominator so I did this:

$$\int \frac{2x-2+2+1}{x^2-2x+2}dx = \int \frac{2x-2}{x^2-2x+2}dx + \int \frac{3}{x^2-2x+2}dx$$

so the first integral is:

$$\int \frac{2x-2}{x^2-2x+2}dx = log|x^2-2x+2| + C$$

But not I have no idea about how to procede with the second integral, I get stucked here:

$$\int \frac{3}{x^2-2x+2}dx = 3\int \frac{1}{x^2-2x+2}dx$$

Should I use partial fractions? There is another way to procede?

Thanks.

Christian Giupponi
  • 687

1 Answers1

2

In denominator, create perfect square by using $$ x^2 -2x +2 = (x-1)^2 + 1 $$ And then replace x-1 by t, or solve directly.

Yash
  • 291