Integration by parts: check my work please :)
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Integration by parts: check my work please :)

[From: ] [author: ] [Date: 11-09-15] [Hit: ]
Where did this one come from? The integral of (4+x)^1/2 is NOT 2x^3/3,and youll have to integrate the vdu term.Watch the algebra as it gets messy.......
∫x*(4+x)^1/2 dx

u=x du=1dx
v=2x^3/3 dv=(4+x)^1/2

∫udv= uv-∫vdu
∫x*(4+x)=(2x^4)/3 - ∫(2x^3)/3 dx
=(2x^4)/3 - 2/3∫x^3 dx
=(2x^4)/3 - (2x^4/12)
=2/3[x^4 - x^4/4)

is this the right answer????

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Take the derivative! And find out that it is the WRONG answer.

"v=2x^3/3 dv=(4+x)^1/2"? Where did this one come from? The integral of (4+x)^1/2 is NOT 2x^3/3, it is
(2/3)(x+4)^3/2

and you'll have to integrate the vdu term.

The final answer is (2/15)(3x-8)(x+4)^3/2

Watch the algebra as it gets messy.
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