Integration question

[tex]\int \frac{1}{x\sqrt{4x + 1}}dx[/tex]
I let [tex]u= \sqrt{4x+1}[/tex], [tex]u^2 = 4x+1[/tex]
So [tex]\frac{1}{2}u du = dx[/tex]

[tex]\int \frac{\frac{1}{2}u}{(\frac{u^2-1}{4})*((\sqrt{u})^2)}du[/tex]
[tex]\int \frac{2u}{(u^2 - 1)*(u)} du[/tex]
[tex]\int \frac{2}{u^2 - 1} du[/tex]
[tex]2\int \frac{1}{u^2 -1}du[/tex]

Can anyone help me anti-differentiate what the integrand is?
 
Last edited:

James R

Science Advisor
Homework Helper
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Isn't it related to the inverse hyperbolic tangent of u?
 
664
0
Try u=sinθ.
 
694
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I'd suggest partial fraction decomposition...
 

TD

Homework Helper
1,020
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Expanding that fractions gives:

[tex]\frac{1}
{{u^2 - 1}} = \frac{1}
{{2\left( {u - 1} \right)}} - \frac{1}
{{2\left( {u + 1} \right)}}[/tex]
 

James R

Science Advisor
Homework Helper
Gold Member
601
15
Partial fractions work. So does u=tanh t.
 

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