How to integrate $\int\frac{-x-1}{(x^2-2x+5)}dx$

Question

How do you integrate $$\int\dfrac{-x-1}{(x^2-2x+5)}dx$$ ?

I would be really grateful for an answer.

Answer 1

The polynomial at denominator has negative discriminant; for this there is a specific technique.

First of all “complete the square”: $x^2-2x+5=x^2-2x+1+4=(x-1)^2+4$. Then make this a sum of squares: $(x-1)^2+4=(x-1)^2+2^2$.

Set $t=(x-1)/2$, so $x=2t+1$ and $dx=2\,dt$, so the integral becomes $$ \int\frac{-2t-1-1}{4t^2+4}2\,dt=-\int\frac{t+1}{t^2+1}\,dt $$ Split this into $$ -\frac{1}{2}\int\frac{2t}{t^2+1}\,dt-\int\frac{1}{t^2+1}\,dt $$

The first integral has at the numerator the derivative of the denominator, so…

The second integral is elementary.

Answer 2

As $\dfrac{d(x^2-2x+5)}{dx}=2x-2$

Write $-x-1=\dfrac{-2x-2}2=\dfrac{-(2x-2)-4}2=-\dfrac12(2x-2)-2$

Now $\int\dfrac{dx}{(x^2-2x+5)}=\int\dfrac{dx}{(x-1)^2+2^2}$