Solving $\int \frac{1}{2x^2+3x+2}dx$ with partial fractions

Question

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?

Answer 1

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 $\frac{2}{\sqrt{7}}\arctan\mathopen{}\left(\frac{4}{\sqrt{7}}x+\frac{3}{\sqrt{7}}\right)\mathclose{}$.

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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.

Answer 2

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:

$$=\frac12\times\frac4{\sqrt{7}}\arctan\frac{u}{\frac{4}{\sqrt{7}}}+c,c\in\mathbb{R}$$

$$=\frac{2}{\sqrt7}\arctan\frac{4u}{\sqrt7}+c$$

$$=\frac{2}{\sqrt7}\arctan\frac{4(x+\frac34)}{\sqrt7}+c$$