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Quick Question

  • Thread starter laker88116
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57
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[tex] \int \frac {1}{x\sqrt{4x+1}}dx[/tex]

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

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

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

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

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

Answers and Replies

663
0
laker88116 said:
[tex] \int \frac {1}{x\sqrt{4x+1}}dx[/tex]

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

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

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

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

I know I have to use partial fractions. But I am not sure where to start. Any help is appreciated.
[tex]\frac{1}{u(u+1)(u-1)}=\frac{A}{u}+\frac{B}{u+1}+\frac{C}{u-1}[/tex]
 
57
0
I know that, but then what do I do, I have to get a common denominator, which is [tex] u(u+1)(u-1) [/tex], but then what?
 
663
0
laker88116 said:
I know that, but then what do I do, I have to get a common denominator, which is [tex] u(u+1)(u-1) [/tex], 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.
 
57
0
Ok, I get A=-1, b=1/2, c=1/2, is that right?
 
663
0
laker88116 said:
Ok, I get A=-1, b=1/2, c=1/2, is that right?
Looks good.
 
57
0
Ok, I think I got it. Thanks.
 
Fermat
Homework Helper
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laker88116 said:
[tex] \int \frac {1}{x\sqrt{4x+1}}dx[/tex]

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

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

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

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

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.

[tex]\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} [/tex]

Substituting gives [tex]\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[/tex]

Now you can do the partial fractions bit.
 

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