# Math Help - integral

1. ## integral

1/16-(x+1)^2

here are my steps;
1/16-x^2-2x-1

1/-x^2-2x+15

then use partial rule.
(-x-5)(x-3)

am i right?

2. Originally Posted by questionboy
1/16-(x+1)^2

here are my steps;
1/16-x^2-2x-1

1/-x^2-2x+15

then use partial rule.
(-x-5)(x-3)

am i right?
$16 - (x+1)^2 = [4 - (x+1)][4 + (x+1)] = (3-x)(5+x) = -(x-3)(x+5)$

3. Aren't they same?

4. Originally Posted by questionboy
Aren't they same?
yes

5. Originally Posted by questionboy
Replying with a Headbang is pointless.

I suspect from skeeter's replies that he was hoping you would now proceed to get the required partial fraction decomposition. You've been taught how to do that, right?