Find the indefinite integral using integration by substitution: Sx^2 sqrt(x+1) dx i figured u=x+1, x=u-1, du=dx S sqrt u  du (2/3)u^(3/2) +C (2/3)(x+1)^(3/2) +C

Expert Answers

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Evaluate `int x^2sqrt(x+1)dx` :

If we use your substitution:

`u=x+1,du=dx,x=u-1` we can rewrite the integral as:

`int(u-1)^2u^(1/2)du`

`=int(u^2-2u+1)u^(1/2)du`

`=int(u^(5/2)-2u^(3/2)+u^(1/2))du`

The integral of a sum is the sum of integrals so:

`=2/7u^(7/2)-4/5u^(5/2)+2/3u^(3/2)+c_1`

Substituting for u we get:

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`=2/7(x+1)^(7/2)-4/5(x+1)^(5/2)+2/3(x+1)^(3/2)+C` ***********************

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This can be rewritten; factor out the common `(x+1)^(3/2)` and a common fraction `2/105` to get:

`=2/105(x+1)^(3/2)[15(x+1)^2-42(x+1)+35]+C`

`=2/105(x+1)^(3/2)[15x^2+30x+15-42x-42+35]+C`

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`=2/105(x+1)^(3/2)[15x^2-12x+8]+C` *************************

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