# Thread: another area under curves

1. ## another area under curves

$y=x^2
y=\frac{2}{x^2+1}
$

intersects at + or - 1
so integral would be $
2\int_{-1}^1 \frac{2}{x^2+1}-x^2dx$

after reducing, I got $2\int_{-1}^1 \frac{1-x^2}{x^2+1}dx$
Don't know how to find anti derivative from here...

I just looked in the back for the answer and it is pi-2/3
is there some sort of trig identity going on....

2. Originally Posted by dorkymichelle
$y=x^2
y=\frac{2}{x^2+1}
$

intersects at + or - 1
so integral would be $
2\int_{-1}^1 \frac{2}{x^2+1}-x^2dx$

after reducing, I got $2\int_{-1}^1 \frac{1-x^2}{x^2+1}dx$
Don't know how to find anti derivative from here...
Personally, I don't know why you have that final line. It's unnecessary and also wrong. Just use the second last line (after fixing an obvious error in it) ....