# Math Help - i dont really understand the last step of this inverse trig integration

1. ## i dont really understand the last step of this inverse trig integration

i got (integral) 1/(u^2 +7/4) du

the answer is [2/sq.rt.7]( arctan (2u/sq.rt.7))

arctan u is the integral of (1/1+u^2)

2. Originally Posted by twostep08
i got (integral) 1/(u^2 +7/4) du

the answer is [2/sq.rt.7]( arctan (2u/sq.rt.7))

arctan u is the integral of (1/1+u^2)
Remember that in general $\int\frac{1}{x^2+a^2}dx=\int\frac{dx}{a^2\left(\fr ac{x^2}{a^2}+1\right)}=\frac{1}{a^2}\int\frac{dx}{ \left(\frac{x}{a}\right)^2+1}$

In this case, your $a^2=\frac{7}{4}$

So you're gonna need to use a substitution $u=\frac{2x}{\sqrt{7}}$.

EDIT: Sorry I used a different variable. I forgot that you're using u instead of x lol.