# Thread: Integration by parts involving complex roots

1. ## Integration by parts involving complex roots

How do I integrate (2x + 3)/(x^2 + 2x + 3)

The denominator has complex roots so I can't break it up to integrate by partial fractions.

2. ## Re: Integration by parts involving complex roots

do some 'jugglery' so that you'll get

$\frac{2x+2}{x^2+2x+3}+\frac{1}{(x+1)^2+2}$

I think now you can use,

$\int \frac{f'(x)}{f(x)} dx = ln(f(x))$and $\int \frac{1}{1+x^2} dx = \tan^{-1}x$

3. ## Re: Integration by parts involving complex roots

Break it up into (2x+2)/(x^2+2x+3)+1/(x^2+2x+3)

For the 1st part numerator is derivative of denominator so integral is ln(x^2+2x+3)

Write 2nd part as 1/(x+1)^2+2 and substitute x+1=root2tanu

4. ## Re: Integration by parts involving complex roots

Originally Posted by necromanzer52
How do I integrate (2x + 3)/(x^2 + 2x + 3)

The denominator has complex roots so I can't break it up to integrate by partial fractions.
Well, you can do exactly that, but this is a better method...

\displaystyle \begin{align*} \int{\frac{2x+3}{x^2+2x+3}\,dx} &= \int{\frac{2x+2}{x^2+2x+3}\,dx} + \int{\frac{1}{x^2+2x+3}\,dx} \\ &= \int{\frac{2x+2}{x^2+2x+3}\,dx} + \int{\frac{1}{x^2 + 2x + 1^2 - 1^2 + 3}\,dx} \\ &= \int{\frac{2x+2}{x^2 + 2x+3}\,dx} + \int{\frac{1}{\left(x + 1\right)^2 + 2}\,dx} \end{align*}

The first integral can be solved by making the substitution \displaystyle \begin{align*} u = x^2 + 2x + 3 \implies du = \left( 2x + 2 \right) dx \end{align*} and the second integral can be solved by making the substitution \displaystyle \begin{align*} x + 1 = \sqrt{2}\tan{\theta}\,d\theta \implies dx = \sec^2{\theta}\,d\theta \end{align*}, the integral becomes

\displaystyle \begin{align*} \int{\frac{2x+2}{x^2+2x+3}\,dx} + \int{\frac{1}{\left(x+1\right)^2+2}\,dx} &= \int{\frac{1}{u}\,du} + \int{\frac{1}{\left(\sqrt{2}\tan{\theta}\right)^2 + 2} \, \sqrt{2}\sec^2{\theta}\,d\theta} \\ &= \int{\frac{1}{u}\,du} + \int{\frac{\sqrt{2}\sec^2{\theta}}{2\tan^2{\theta} + 2}\,d\theta} \\ &= \int{\frac{1}{u}\,du} + \int{\frac{\sqrt{2}\sec^2{\theta}}{2\left(\tan^2{ \theta} + 1\right)}\,d\theta} \\ &= \int{\frac{1}{u}\,du} + \int{\frac{\sqrt{2}\sec^2{\theta}}{2\sec^2{\theta} }\,d\theta} \\ &= \int{\frac{1}{u}\,du} + \int{\frac{\sqrt{2}}{2}\,d\theta} \\ &= \ln{|u|} + \frac{\sqrt{2}}{2}\theta + C \\ &= \ln{\left| x^2 + 2x + 3 \right|} + \frac{\sqrt{2}}{2}\arctan{ \left[ \frac{ \sqrt{2} \left( x+1 \right) }{2} \right] } + C \end{align*}