# 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

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

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...

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

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:

-----------------------------------------------------------------

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

------------------------------------------------------------------

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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