Calculus 1 Integration Problem

[kikko]

Homework Statement

Calc I problem, so need to solve by calc 1 methods.

∫ (x-1)(x+1)^11 dx

Homework Equations

The Attempt at a Solution

∫ (x-1)(x+1)^11 dx

U = x+1
dU = dx
x-1 = x+1-2 = U-2

=∫ (U-2)U^11 dU

=∫ U^12 dU - 2∫ U^11 dU

=(1/13)U^13 - (1/6)U^12

=(1/13)(x+1)^13 - (1/6)(X+1)^12

∫ (x-1)(x+1)^11 dx = (1/13)(x+1)^13 - (1/6)(X+1)^12
The course ended already, this was the final test problem (on the test that ended already). I didn't understand how to do this one. I can check my answer and see it's wrong. My teacher won't be in until fall semester starts, so I can't ask him until then. I was already at full time, so Calc I was basically a free class I retook because it could boost GPA without being time consuming.

 

[XtremePhysX]

\int (x-1)(x+1)^{11} dx \\ = \int x(x+1)^{11}-(x+1)^{11} dx \\ = \int [(x+1)-1](x+1)^{11}-(x+1)^{11} dx \\ = \int (x+1)^{12}-2(x+1)^{11} dx \\ = \frac{(x+1)^{13}}{13}-\frac{(x+1)^{12}}{6} +C

 

[kikko]

That seems to be the answer I got. Forgot all those constants on my answers >_<. When I take the derivative of that I get:

(x+1)^12 - 2(x+1)^11 =/= (x-1)(x+1)^11

I might be missing something here.

 

[Villyer]

That seems to be the answer I got. Forgot all those constants on my answers >_<. When I take the derivative of that I get:

(x+1)^12 - 2(x+1)^11 =/= (x-1)(x+1)^11

I might be missing something here.

But it is, isn't it?

$(x+1)^{12} - 2(x+1)^{11}$
$(x+1)(x+1)^{11} - 2(x+1)^{11}$
$(x+1)^{11}(x+1-2)$
$(x-1)(x+1)^{11}$