Solving a Complex Integral with Partial Fractions

[laker88116]

\int \frac {1}{x\sqrt{4x+1}}dx

Here's what I have done so far on this problem

I let u= \sqrt{4x+1}, so then u^2=4x+1, du= \frac {2dx}{u} and x= \frac {u^2-1}{4}

Substituting, I get \int \frac {1}{(\frac{u^2-1}{4})u}du

Then moving stuff around, I get 4 \int \frac {du}{u(u+1)(u-1)}

I know I have to use partial fractions. But I am not sure where to start. Any help is appreciated.

 

[amcavoy]

\int \frac {1}{x\sqrt{4x+1}}dx

Here's what I have done so far on this problem

I let u= \sqrt{4x+1}, so then u^2=4x+1, du= \frac {2dx}{u} and x= \frac {u^2-1}{4}

Substituting, I get \int \frac {1}{(\frac{u^2-1}{4})u}du

Then moving stuff around, I get 4 \int \frac {du}{u(u+1)(u-1)}

I know I have to use partial fractions. But I am not sure where to start. Any help is appreciated.

\frac{1}{u(u+1)(u-1)}=\frac{A}{u}+\frac{B}{u+1}+\frac{C}{u-1}

 

[laker88116]

I know that, but then what do I do, I have to get a common denominator, which is u(u+1)(u-1), but then what?

 

[amcavoy]

I know that, but then what do I do, I have to get a common denominator, which is u(u+1)(u-1), but then what?

Multiply both sides by u(u+1)(u-1) so that the left side is 1. Simplify to get A, B, and C.

 

[laker88116]

Ok, I get A=-1, b=1/2, c=1/2, is that right?

 

[amcavoy]

Ok, I get A=-1, b=1/2, c=1/2, is that right?

Looks good.

 

[laker88116]

Ok, I think I got it. Thanks.

 

[Fermat]

\int \frac {1}{x\sqrt{4x+1}}dx

Here's what I have done so far on this problem

I let u= \sqrt{4x+1}, so then u^2=4x+1, du= \frac {2dx}{u} and x= \frac {u^2-1}{4}

Substituting, I get \int \frac {1}{(\frac{u^2-1}{4})u}du

Then moving stuff around, I get 4 \int \frac {du}{u(u+1)(u-1)}

I know I have to use partial fractions. But I am not sure where to start. Any help is appreciated.

Just a little bit missed out. When substituting into the integrand written in terms of x, you forgot to replace the dx with (u/2)du.

\mbox{let}\ u= \sqrt{4x+1}\ \mbox{, so then}\ u^2=4x+1\ \mbox{,}\ du= \frac {2dx}{u}\ \rightarrow dx = \frac{u}{2}\ du \mbox{, and}\ x= \frac {u^2-1}{4}

Substituting gives \int \frac {1}{x\sqrt{4x+1}}dx = \int \frac {1}{(\frac{u^2-1}{4})u}.\frac{u}{2}\ du = \int \frac{2}{u^2 - 1}\ du

Now you can do the partial fractions bit.