# Tough Integral

• Apr 19th 2008, 04:46 PM
Thomas
Tough Integral
I've tried everything I can think of and I'm not getting it. I'm studying for a final right now, so if there's a trick to doing this question I figure I should know it. :)

$\displaystyle \int \frac{x + 1}{\sqrt{4x-x^2}} dx$

I've tried trig substitution, integration by parts, normal substitution, etc.

Any ideas?
• Apr 19th 2008, 04:58 PM
TheEmptySet
Quote:

Originally Posted by Thomas
I've tried everything I can think of and I'm not getting it. I'm studying for a final right now, so if there's a trick to doing this question I figure I should know it. :)

$\displaystyle \int \frac{x + 1}{\sqrt{4x-x^2}} dx$

I've tried trig substitution, integration by parts, normal substitution, etc.

Any ideas?

try this

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

$\displaystyle -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}}dx+\int \frac{3}{\sqrt{4-(x-2)^2}}dx$

Now make u subs in each integrals :D
• Apr 19th 2008, 05:03 PM
Mathstud28
Quote:

Originally Posted by TheEmptySet
try this

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

$\displaystyle -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}}dx+\int \frac{3}{\sqrt{4-(x-2)^2}}dx$

Now make u subs in each integrals :D

Yeah...but its not u-substitution for the last one it is $\displaystyle 4arcsin\bigg(\frac{x-2}{2}\bigg)+C$
• Apr 19th 2008, 05:07 PM
Mathstud28
Quote:

Originally Posted by TheEmptySet
try this

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

$\displaystyle -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}}dx+\int \frac{3}{\sqrt{4-(x-2)^2}}dx$

Now make u subs in each integrals :D

Also you can make a simpler integral usign a different partial fractions decomposition
• Apr 19th 2008, 05:13 PM
TheEmptySet
Quote:

Originally Posted by Mathstud28
Yeah...but its not u-substitution for the last one it is $\displaystyle 4arcsin\bigg(\frac{x-2}{2}\bigg)+C$

It can be 2u=x-2 du=dx

$\displaystyle \frac{3}{\sqrt{4-(x-2)^2}}dx=3\int \frac{1}{\sqrt{4-4u^2}}du=\frac{3}{2}\int\frac{1}{sqrt{1-u^2}}du$

Now sub $\displaystyle u=\sin(\theta), du=\cos(\theta)$

I guess you could do it all in one step with $\displaystyle 2\sin(\theta)=x-2$

but I usually do mine one at a time.(We all can't be inte-killer's) :)
• Apr 19th 2008, 05:17 PM
TheEmptySet
Quote:

Originally Posted by Mathstud28
Also you can make a simpler integral usign a different partial fractions decomposition

I would like to see the decomp it you can please!

I don't think it can be done.

I thought it only worked with rational functions.
• Apr 19th 2008, 05:26 PM
Thomas
Thanks for the help, got the answer. I don't think I've ever used, or seen, that technique before.

But, this was supposed to be an overly hard final to make up for an easy midterm, so no wonder the questions are so hard. :p