# Thread: Partial fraction integration help!

1. ## Partial fraction integration help!

Alright, these have been giving me heck all day. I have two short ones left. It's been a long day full of beating up on these monstrous ogres of problems ( Something like a 6x5 matrix for one of them ) and I'm glad to say these should be awesomely simple.

First, The integral of 1/(x(x^4+1) dx. My book has a hint: "Multiply by x^3/x^3" on this problem. I broke it out into A/X + (B*X+C)/(x^4+1) = 1/(x(x^4+1)). Then I eliminated fractions. It's pretty easy to see A=1.

Then I was promptly lost in the woods.

Second one that looks a bit more intimidating is the integral of 1/(x^(3/2) - x^(1/2)).

I started by letting U=x^(1/2) and DU=1/2(x^(1/2)) I ran through this one to the end, and ended up with LN(|U-1|/|u|)+C but that denominator should have been |u+1|.

Thanks for the help! I really appreciate what you guys do here and I am linking this site to friends.

2. Why didn't you do what was suggested?

$\displaystyle \displaystyle \int{\frac{1}{x(x^4 + 1)}\,dx} = \int{\frac{x^3}{x^4(x^4 + 1)}\,dx}$.

Now make a very simple substitution...

3. Simply let $\displaystyle t = \frac{1}{x}$, then your integral:

\displaystyle \begin{aligned}& I = -\int\frac{1/t^2}{1/t\left(\frac{1}{t^4}+1}\right)}\;{dt} = -\int\frac{t^3}{1+t^4}\;{dt} = -\frac{1}{4}\int\frac{4t^3}{1+t^3}\;{dt} \\ & = -\frac{1}{4}\int\frac{(1+t^4)'}{1+t^4}\;{dt} = -\frac{1}{4}\ln\left(1+t^4\right)+k = -\frac{1}{4}\ln\left(1+\frac{1}{x^4}\right)+k. \end{aligned}

4. Originally Posted by TheCoffeeMachine
Simply let $\displaystyle t = \frac{1}{x}$, then your integral:

\displaystyle \begin{aligned}& I = -\int\frac{1/t^2}{1/t\left(\frac{1}{t^4}+1}\right)}\;{dt} = -\int\frac{t^3}{1+t^4}\;{dt} = -\frac{1}{4}\int\frac{4t^3}{1+t^3}\;{dt} \\ & = -\frac{1}{4}\int\frac{(1+t^4)'}{1+t^4}\;{dt} = -\frac{1}{4}\ln\left(1+t^4\right)+k = -\frac{1}{4}\ln\left(1+\frac{1}{x^4}\right)+k. \end{aligned}
Of course, this goes against what the OP was told to do in the problem, i.e. to multiply top and bottom by $\displaystyle \displaystyle x^3$...

The substitution made after doing this is also much easier...

5. ROFL.

Okay, I assumed they meant "Multiply after doing something to it.".

I think my problem there may be of the T=pi/2 AM ID(10T) variety.

I knew it was something easy. I'm almost afraid to ask about the other problem but I do need to practice!

6. Originally Posted by Wolvenmoon
I knew it was something easy. I'm almost afraid to ask about the other problem but I do need to practice!
Let $\displaystyle t = \sqrt{x}$ so that $\displaystyle dx = 2\sqrt{x} \;{dt}$ you will end up with:

\displaystyle \displaystyle \begin{aligned} I & = 2\int\frac{1}{t^2-1}\;{dt} = \int\frac{(t+1)-(t-1)}{(t-1)(t+1)}\;{dt} = \int\frac{1}{t-1}-\frac{1}{t+1}\;{dt} \\& = \ln\left(t-1\right)-\ln\left(t+1\right)+k = \ln\left|\frac{t-1}{t+1}\right|+k = \ln\left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|+k. \end{aligned}

7. Originally Posted by Prove It
Of course, this goes against what the OP was told to do in the problem, i.e. to multiply top and bottom by $\displaystyle \displaystyle x^3$...
Of course, I was just pointing out another way that it could be done.
The substitution made after doing this is also much easier.
To be fair, I think both substitutions are very easy.