
Setting

, you have

. Then the integral becomes




Now,

in general. But since we want our substitution

to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means

, which implies that

, or equivalently that

. Over this domain,

, so

.
Long story short, this allows us to go from

to


Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

Then integrate term-by-term to get


Now undo the substitution to get the antiderivative back in terms of

.

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to
19191020202929293994494944
=3(3z)2-3
=3(9z2)-3
=27z^2-3
=3(9z^2-1)
=3(3z+1)(3z-1)
(2•7)-5
14-5= 9
So the answer is 9
Answer:
3/10 or three-tenths
Step-by-step explanation:
Hope this helps=D