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Posted

yes, I tried setting u=(x^2+1)^(1/2) and dv=dx

but I am not getting any log in the answer which means wrong.

using the parts on sec(theta)^3 gets me nowhere, I change to (tan(x)^2+1)sec(x) and then set u=tan(x)^2 and dv=sec(x) dx which lead me to very complicated. I know I did this before but I don't remember how I did it.

Posted

Use trig substitution such that [math]x[/math] = [math]\tan{\theta}[/math]

Then it will be easy.

 

 

[edit]

When you get to [math]\int{\sec{\theta}^3 dx}[/math], use parts with [math]u = \sec{\theta} [/math] and [math] dv = \sec{\theta}^2 dx [/math] and then use trig identities to get the [math]\int{\sec{\theta}^3 dx}[/math] again. Once you have that, move it to the left side and you're done. (Sub in x)

[/edit]

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