Integrate In anyway possible.

**EliteAndoy** · February 12th 2013, 10:38 PM

Hi everyone! Today we are asked in class to solve an integral in anyway that we know. The given integral is $\int \frac{x^4}{x-1}dx$ . She hints that it is possible to solve it using integration by parts, and I see that I might have to use tablature method to find the integral since I have to reduce $x^4$ to zero power. But I said, screw it! So I attempted to do it by doing somewhat partial fraction. So here's what I did:

1)I did a long division so that:

$\frac{x^4}{x-1}=x^3+x^2+x+1+\frac{1}{x-1}$

2)Turns out I don't have to solve for a partial fraction so:

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

3)So that if we integrate it, it becomes:

$\int \frac{x^4}{x-1}dx=\frac{x^4}{4}+\frac{x^3}{3}+\frac{x^2}{2}+x+l n\left |x-1\right |+c$

Is that solution good enough? Not really sure about it. Anyone got any ideas to evaluate the integral more efficiently? Thanks a lot in advance!

**Prove It** · February 12th 2013, 11:33 PM

Originally Posted by EliteAndoy

Hi everyone! Today we are asked in class to solve an integral in anyway that we know. The given integral is $\int \frac{x^4}{x-1}dx$ . She hints that it is possible to solve it using integration by parts, and I see that I might have to use tablature method to find the integral since I have to reduce $x^4$ to zero power. But I said, screw it! So I attempted to do it by doing somewhat partial fraction. So here's what I did:

1)I did a long division so that:

$\frac{x^4}{x-1}=x^3+x^2+x+1+\frac{1}{x-1}$

2)Turns out I don't have to solve for a partial fraction so:

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

3)So that if we integrate it, it becomes:

$\int \frac{x^4}{x-1}dx=\frac{x^4}{4}+\frac{x^3}{3}+\frac{x^2}{2}+x+l n\left |x-1\right |+c$

Is that solution good enough? Not really sure about it. Anyone got any ideas to evaluate the integral more efficiently? Thanks a lot in advance!

Let $\displaystyle \begin{align*} u = x - 1 \implies du = dx \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{\frac{x^4}{x - 1}\,dx} &= \int{\frac{\left( u + 1 \right)^4}{u}\,du} \\ &= \int{\frac{u^4 + 4u^3 + 6u^2 + 4u + 1}{u}\,du} \\ &= \int{u^3 + 4u^2 + 6u + 4 + \frac{1}{u} \,du} \\ &= \frac{u^4}{4} + \frac{4u^3}{3} + 3u^2 + 4u + \ln{|u|} + C \\ &= \frac{ \left( x - 1 \right)^4 }{4} + \frac{ 4\left( x - 1 \right)^3}{3} + 3\left( x - 1 \right)^2 + 4 \left( x- 1 \right) + \ln{ \left| x - 1 \right| } + C \end{align*}$

**EliteAndoy** · February 12th 2013, 11:40 PM

That also works, but still I'm not certain if my solution is correct. What do you think?

**MINOANMAN** · February 12th 2013, 11:56 PM

yes your solution is correct
the best way to solve it is the one discribed by prove it.

I suggest you study more the integration ...
MINOAS

**ibdutt** · February 13th 2013, 02:28 AM

In such cases i have found that partial fraction approach is always better.

**EliteAndoy** · February 13th 2013, 07:50 AM

Thanks everyone!

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