- #1

Mentallic

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## Homework Statement

[tex]\int{\frac{xsec^2x}{tanx+\sqrt{3}}}dx[/tex]

## The Attempt at a Solution

I let [tex]tanx+\sqrt{3}=u[/tex]

Then, [tex]\frac{du}{dx}=sec^2x , dx=\frac{du}{sec^2x}[/tex]

Substituting into the integral: [tex]\int{\frac{x}{u}du[/tex]

But now I don't know what to do. That little x is really screwing me over on this... Maybe a whole other approach needs to be taken?