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

Please show your working and thanks for the help.

Also, please refer to the techniques of integration you are using.

Re: Indefinite integral of (x+(1/x))^4

**$\displaystyle \int \left(x+\frac{1}{x}\right)^4 \, dx$ = $\displaystyle \int \left(6+\frac{1}{x^4}+\frac{4}{x^2}+4 x^2+x^4\right) \, dx$**

Re: Indefinite integral of (x+(1/x))^4

Please show your working throughout and please simplify

Re: Indefinite integral of (x+(1/x))^4

We are given:

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

Using the binomial theorem on the integrand, we find:

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

After simplification:

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

So then we have:

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

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

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

So we have:

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