How Do You Integrate 9x³/sqrt(1+x²) with a Trigonometric Substitution?

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Homework Statement



integrate this

9x³/sqrt(1+x²)

I know I am suppose to substitute using: x = tan x
dx = sec²x

but that's as far as i could go. its the 9x³ that confuses me on what i should do.
thank you for your help in advance.
 
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You should not be substituting x = tan x. You should be substituting x = tan u. Then, dx = sec2u du. So substitute tan u everywhere there is a x, and sec2u du for dx to get a complete integral in u.
 
x= \sinh u as a substitution should do the trick.
 
bigubau said:
x= \sinh u as a substitution should do the trick.

You don't need trig at all. u=1+x^2 will work fine. Try it.
 
thanks guys for all the help really appreciate it
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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