Asked by Molly
Find the integral, using the method of substitution where necessary.
63.
integrate x^2*Sqrt[x+1] dx
What I have done: I have set the (x+1) = u
but I feel this is the wrong way to do this problem.
63.
integrate x^2*Sqrt[x+1] dx
What I have done: I have set the (x+1) = u
but I feel this is the wrong way to do this problem.
Answers
Answered by
Damon
That is a fine substitution to make
u = x + 1
dx = du
x^2 = u^2 - 2u+1
so we have
integral (u^2-2u+1)(u^.5) du
that is integral of u^2.5 du
- integral of 2u^1.5 du
+ integral of u^.5 du
in general
integral of u^n du = [u^(n+1)]/(n+1)
u = x + 1
dx = du
x^2 = u^2 - 2u+1
so we have
integral (u^2-2u+1)(u^.5) du
that is integral of u^2.5 du
- integral of 2u^1.5 du
+ integral of u^.5 du
in general
integral of u^n du = [u^(n+1)]/(n+1)
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