I have an integral with an irreducible quadratic. I'm supposed to use partial fractions to solve the integral.
The original integral: $\int\dfrac{x^2}{(x-1)(x^2+4x+5)}dx$. I used partial fractions to break the integral up and got the following result after simplifying the resulting fractions:
$$\dfrac{1}{10}\int\dfrac{1}{x-1}dx + \dfrac{1}{10}\int\dfrac{9x+5}{x^2+4x+5}dx$$
I know how to take care of the $\dfrac{1}{x-1}$ part but I'm not sure how to procede with the second integral that has $\dfrac{9x+5}{x^2+4x+5}$.
I tried substituting the denominator where $u = x^2+4x+5$ and $du = (2x+4)dx$ but I'm not sure what to do with the $2x+4$ that is left over from taking the derivative. Can anyone help?
Complete the square in the denominator to write
$$\int \frac{9x + 5}{(x + 2)^2 + 1} dx$$
Now in the numerator, replace $x$ with $x + 2$, and write
$$9x + 5 = 9(x + 2) - 13$$
Separating the two terms, we find
$$9 \int\frac{x + 2}{(x + 2)^2 + 1} dx - 13 \int \frac{1}{(x + 2)^2 + 1} dx$$
Both of these integrals can be handled by substitutions, leading to a natural logarithm and an inverse tangent term.
In general, a good goal is to write terms involving
$$\int \frac{t}{t^2 + 1} dt = \frac 1 2 \ln{(t^2 + 1)} + C$$
and
$$\int \frac{1}{t^2 + 1} dt = \arctan(t) + C$$