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I don't know how to solve these. How do you evaluate the integral of [tex]\frac{3dx}{\sqrt{3+X^2}}[/tex]? I know you have to set x=atan[tex]\theta[/tex].

our a is [tex]\sqrt{3}[/tex] so x =[tex]\sqrt{3}[/tex]tan[tex]\theta[/tex]. That means

dx=[tex]\sqrt{3}[/tex]sec^{2}[tex]\theta[/tex]d[tex]\theta[/tex].

I also made a right triangle using the information that tan[tex]\theta[/tex]=X/sqrt(3)

So the two legs of the triangle are X and sqrt(3) and the hypotenuse is sqrt{X^2+3

I try to solve using substitution but I can never get the right answer.

The answer is supposed to be 3*ln(X + sqrt(3+X^2) + C

I can't seem to get tags to work properly so just bear with the poor formatting.

I have a test on this stuff tomorrow and this is a review problem. Sadly, my professor did not give us solutions; just answers.

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# Integrals involving trig substitution

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