- #1

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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.