# Indefinite integral of (x+(1/x))^4

• Sep 13th 2012, 11:00 PM
fbarawr
Indefinite integral of (x+(1/x))^4
Hello, I'm Fahad and I'm new here
I'm here to get help with simple math problems for my IGCSES
Plese help me.
The following math is from the binomial expansions part of my syllabus:

⌠ (x+(1/x))^4

Also, please refer to the techniques of integration you are using.
• Sep 13th 2012, 11:06 PM
MaxJasper
Re: Indefinite integral of (x+(1/x))^4
$\int \left(x+\frac{1}{x}\right)^4 \, dx$ = $\int \left(6+\frac{1}{x^4}+\frac{4}{x^2}+4 x^2+x^4\right) \, dx$
• Sep 13th 2012, 11:08 PM
fbarawr
Re: Indefinite integral of (x+(1/x))^4
• Sep 13th 2012, 11:12 PM
MarkFL
Re: Indefinite integral of (x+(1/x))^4
We are given:

$\int \left(x+\frac{1}{x} \right)^4\,dx$

Using the binomial theorem on the integrand, we find:

$\left(x+\frac{1}{x} \right)^4=\left(x+x^{-1} \right)^4=\sum_{k=0}^4{4 \choose k}x^{4-k}x^{-k}$

After simplification:

$x^4+4x^2+6+4x^{-2}+x^{-4}$

So then we have:

$I=\int x^4+4x^2+6+4x^{-2}+x^{-4}\,dx$

We will integrate term by term, using the power rule for integration $\int k\cdot x^n\,dx=\frac{k}{n+1}x^{n+1}+C$

where k is a constant and $n\ne-1$

So we have:

$I=\frac{1}{5}x^5+\frac{4}{3}x^3+6x-4x^{-1}-\frac{1}{3}x^{-3}+C$