Integration for two problems

[alimarie333]

I am extremely stuck on two integration problems that I have been trying to approach the last 2 hours. If anyone can help me on these I would be grateful! Thank u

1. I tried using integration by parts but it was getting so messy:

integral of (x^4 + 1)/(x(x^2+1)^2) dx

2. integral of 1/(1+ cube root of x) dx from 0 to 1.
I know I can set u= cube root of x
For this one, I know how to integrate it when the denominator is square root of x instead of cube root of 3. If this were the case then u=square root of x, dx is 2udu. The answer would then be 2-2ln2

Thank u again!

 

[mark07]

1) You need to use partial fractions on this one:

\(\displaystyle \L \frac{x^4 + 1}{x(x^2+1)^2} = \frac{A}{x} + \frac{Bx+C}{x^2+1} +\frac{ Dx+E}{(x^2+1)^2}\)

and from this you need to find A, B, C, D, E.

I get,

\(\displaystyle \L A=1, B=C=0, D=-2, E=0\)

After that, integration is easier, since

\(\displaystyle \L \frac{x^4 + 1}{x(x^2+1)^2} = \frac{1}{x} - \frac{ 2x}{(x^2+1)^2}\)

and

\(\displaystyle \L \int \frac{x^4 + 1}{x(x^2+1)^2} dx = \int \frac{1}{x} dx - \int \frac{ 2x}{(x^2+1)^2} dx\)

First integral is just \(\displaystyle \ln x\) and for the second integral, use u=\(\displaystyle x^2 +1\). I hope that helps.

 

[mark07]

2) Letting \(\displaystyle x=u^3\), you get \(\displaystyle dx = 3u^2 du\), and your integral becomes

\(\displaystyle \L \int \frac{3u^2 du}{1+u}\)

To do this integral, again remember what you learned about rational functions. What happens when the degree of the numerator is bigger than the degree of the numerator?

Looooong divisionnnnn?...

Let me know if you can't finish the rest.