Thread: Integration by Parts

1. Integration by Parts

I have to integrate x times the square root of (x+3) and tried it using integration by parts. If u=x, du=dx, and v=the integration of the square root of (x+3), which is 2/3 times (x+3) raised to the 3/2 power, right?

So, the answer works out to 2/3x times (x+3)^3/2 minus 2/3 times (x+3)^3/2, right?

Also, is the conventional way to solve this integration by parts, or by substitution? I want to know because the answer that you get by u-sub looks a lot different from my answer, and I need to know in case on a multiple choice exam, the test-maker has decided to solve it using u-sub. I thought that you generally don't use u-sub unless you have the derivative of the function (in this case, we have an extra x, which seems to me a cue to not use u-sub).

2. Originally Posted by jaijay32
I have to integrate x times the square root of (x+3) and tried it using integration by parts. If u=x, du=dx, and v=the integration of the square root of (x+3), which is 2/3 times (x+3) raised to the 3/2 power, right?

So, the answer works out to 2/3x times (x+3)^3/2 minus 2/3 times (x+3)^3/2, right?

Also, is the conventional way to solve this integration by parts, or by substitution? I want to know because the answer that you get by u-sub looks a lot different from my answer, and I need to know in case on a multiple choice exam, the test-maker has decided to solve it using u-sub. I thought that you generally don't use u-sub unless you have the derivative of the function (in this case, we have an extra x, which seems to me a cue to not use u-sub).
parts is not necessary ...

$\int x\sqrt{x+3} \, dx$

$u = x+3$

$x = u-3$

$du = dx$

$\int (u-3)\sqrt{u} \, du$

$\int u^{\frac{3}{2}} - 3u^{\frac{1}{2}} \, du$

finish