# Thread: Integrate with Improper Fractions

1. ## Integrate with Improper Fractions

How do you integrate:

$
\int \dfrac {1}{x^3-8}\,dx
$

I got it down to this but i totally stuck on how to integrate one of the improper fraction...

$
\dfrac{-1}{12}\int \dfrac {x+4}{x^2+2x+4}\,dx + \dfrac{1}{2}ln|x-2|
$

What do I do.....??

2. Originally Posted by polymerase
How do you integrate:

$
\int \dfrac {1}{x^3-8}\,dx
$

I got it down to this but i totally stuck on how to integrate one of the improper fraction...

$
\dfrac{-1}{12}\int \dfrac {x+4}{x^2+2x+4}\,dx + \dfrac{1}{2}ln|x-2|
$

What do I do.....??
$\int \frac {x + 4}{x^2 + 2x + 4}~dx = \int \frac {x + 1 + 3}{x^2 + 2x + 4}~dx$

$= \int \frac {x + 1}{x^2 + 2x + 4}~dx + \int \frac {3}{x^2 + 2x + 4}~dx$

Now, for the first integral, use the substitution $u = x^2 + 2x + 4$

for the second integral, complete the square of the denominator and simplify so you can get the arctan antiderivative (recall $\int \frac {1}{x^2 + 1}~dx = \arctan x + C$, there is a formula for $\int \frac {1}{u^2 + a^2}~dx$, but I think practicing the algebraic manipulation would be good for you. then look at the formula)

3. $\displaystyle -\frac{1}{12}\int\frac{x+4}{x^2+2x+4}dx=-\frac{1}{24}\int\frac{2x+2+6}{x^2+2x+4}dx=$
$\displaystyle=-\frac{1}{24}\int\frac{2x+2}{x^2+2x+4}dx-\frac{1}{4}\int\frac{1}{(x+1)^2+3}dx=$
$\displaystyle=-\frac{1}{24}\ln(x^2+2x+4)-\frac{1}{4\sqrt{3}}\arctan\frac{x+1}{\sqrt{3}}+C$