How can I evaluate this integral by partial decomposition?

Question

How can I evaluate this integral, if the denominator has a quadratic factors e.i-($b^{2}-4ac<0$) $$\int\frac{xdx}{x^{2}+6x+13}$$ by partial decomposition?

Answer 1

$$\int\frac{xdx}{x^{2}+6x+13}$$

$$=\frac{1}{2}\int\frac{(2x+6)dx}{x^{2}+6x+13}-\int\frac{3dx}{x^{2}+6x+13}$$

The first term in the integral above has numerator of integrand as derivative of its denominator. So integral can be evaluated by substituting denominator as a variable.

We now need to find :

$$\int\frac{3dx}{x^{2}+6x+13}$$

The general trick is to use completing squares if the quadratic equation has no real roots. So we can write the integral as :

$$\int\frac{3dx}{(x+3)^{2}+2^2}$$

Now substitute $x+3=t$ and then you get :

$$\int\frac{3dt}{t^{2}+2^2}$$

Now complete this using standard integral of inverse tangent function.

Answer 2

Hint:

As mentioned by Anurag, completing the square, you obtain the integral $$\int\frac{x\,\mathrm dx}{(x+3)^2+4}=\int\frac{(x+3)\,\mathrm dx}{(x+3)^2+4}-3\int\frac{\mathrm dx}{(x+3)^2+4}.$$ I'll add that everyone should know the formula $$\int\frac{\mathrm dx}{x^2+a^2}=\frac1a\arctan\Bigl(\frac xa\Bigr).$$