# Integrate. What is the next step ?

## Apr 12, 2018

${x}^{3} / 3 + \frac{35}{12} \ln | x - 2 | + \frac{61}{24} \ln \left({x}^{2} + 2 x + 4\right) + \frac{1}{4 \sqrt{3}} {\tan}^{-} 1 \left(\frac{x + 1}{\sqrt{3}}\right) + C$

#### Explanation:

You have done almost all the heavy lifting here! The first of the two integrals is really simple,

$\frac{35}{12} \int \frac{1}{x - 2} \mathrm{dx} = \frac{35}{12} \ln | x - 2 |$

so I am guessing you need help with the second:

$\frac{1}{12} \int \frac{61 x + 64}{{x}^{2} + 2 x + 4} \mathrm{dx}$

For this, note that

$d \left({x}^{2} + 2 x + 4\right) = \left(2 x + 2\right) \mathrm{dx} = 2 \left(x + 1\right) \mathrm{dx}$

so that it makes sense to split up the numerator into

$61 x + 64 = 61 \left(x + 1\right) + 3$

Then

$\frac{1}{12} \int \frac{61 x + 64}{{x}^{2} + 2 x + 4} \mathrm{dx} = \frac{61}{12} \int \frac{x + 1}{{x}^{2} + 2 x + 4} \mathrm{dx} + \frac{1}{12} \int \frac{3}{{x}^{2} + 2 x + 4} \mathrm{dx}$

The first of these is

$\frac{61}{24} \int \frac{2 \left(x + 1\right) \mathrm{dx}}{{x}^{2} + 2 x + 4} = \frac{61}{24} \ln \left({x}^{2} + 2 x + 4\right)$

while the second is

$\frac{1}{4} \int \frac{\mathrm{dx}}{{x}^{2} + 2 x + 4} = \frac{1}{4} \int \frac{\mathrm{dx}}{{\left(x + 1\right)}^{2} + 3} = \frac{1}{4 \sqrt{3}} {\tan}^{-} 1 \left(\frac{x + 1}{\sqrt{3}}\right)$

Adding all the terms (including the $\int {x}^{2} \mathrm{dx}$) we get the integral :

${x}^{3} / 3 + \frac{35}{12} \ln | x - 2 | + \frac{61}{24} \ln \left({x}^{2} + 2 x + 4\right) + \frac{1}{4 \sqrt{3}} {\tan}^{-} 1 \left(\frac{x + 1}{\sqrt{3}}\right) + C$