Solving $\int \frac{1}{2x^2+3x+2}dx$ with partial fractions I'm learning about integrals, in particular, the partial fractions method.
It appears the first step is to factorize the denominator, and depending on the factors you'll take a different approach. Well, I am having problems with this exercise:
$$\int \frac{1}{2x^2+3x+2}dx$$
Yeah well, I don't think this denominator can be factored anymore. So... yeah. How can I work with this problem?
 A: With an irreducible quadratic denominator, complete the square and substitute to get an integral like $\int\frac{1}{u^2+1}du$, which has $\arctan(u)$ as its antiderivative.
Here, you have $$\frac87\int\frac{1}{\left(\frac{4}{\sqrt{7}}x+\frac{3}{\sqrt{7}}\right)^2+1}\,dx$$ so you can substitute $u=\frac{4}{\sqrt7}x+\frac{3}{\sqrt{7}}$, resulting in an antiderivative of $\frac{2}{\sqrt{7}}\arctan\mathopen{}\left(\frac{4}{\sqrt{7}}x+\frac{3}{\sqrt{7}}\right)\mathclose{}$.

Since you are studying partial fraction reduction in general, you may also see integrands that have an irreducible quadratic denominator like this one, but have a linear (not constant) numerator. For such integrals, you can break them up into the sum of two integrals, one of which has a constant numerator, and the other has some scalar multiple of the denominator's derivative. This latter integral's antiderivative comes from a substituion where $u$ is the full denominator.
A: You need to simplify your fractiondown into the form $\frac{1}{u^2+1}$ which integrates to $\arctan u$ or into the form $\frac{1}{u^2+a^2}$ which integrates to $\frac1a \arctan\frac{x}{a}$. (Depends on what you have remembered/on your cheat sheet.)
$$\int\frac{1}{2x^2+3x+2}dx$$
$$=\int\frac{1}{2(x^2+\frac32x+1)}dx$$
$$=\frac12\int\frac{1}{x+2\cdot\frac34x+\frac9{16}-\frac9{16}+1}dx$$
$$=\frac12\int\frac{1}{(x+\frac34)^2+\frac7{16}}dx$$
Let $u=x+\frac34$, $du=dx$
$$=\frac12\int\frac{1}{u^2+\frac7{16}}du$$
Here we can take $a^2=\frac7{16}$ from above to get:
$$=\frac12\times\frac4{\sqrt{7}}\arctan\frac{u}{\frac{4}{\sqrt{7}}}+c,c\in\mathbb{R}$$
$$=\frac{2}{\sqrt7}\arctan\frac{4u}{\sqrt7}+c$$
$$=\frac{2}{\sqrt7}\arctan\frac{4(x+\frac34)}{\sqrt7}+c$$
