how to you solve this?

Question

Integral [(x cubed) (e to the x squared) / (x squared + 1) squared] dx

Answer 1

u = x² + 1
x² = u - 1
du = 2x dx
x dx = 1/2 du

∫ x³e^(x²)/(x² + 1)² dx =

∫ x²e^(x²)/(x² + 1)² xdx =

1/2∫ (u - 1)e^(u - 1)/u² du =

1/2[∫ e^(u - 1)/u du - ∫ e^(u - 1)/u² du] =

1/2[∫ e^u e^(-1)/u du - ∫ e^u e^(-1)/u² du] =

1/(2e)[∫ e^u/u du - ∫ e^u/u² du] =

Doing integration by parts on the second integral:

∫ e^u/u² du = ∫ uˉ²e^u du

r = e^u
dr = e^u du

ds = uˉ² du
s = -uˉ¹

So....

∫ uˉ²e^u du = -uˉ¹e^u + ∫uˉ¹e^u du

Resubstituting:

1/(2e)(∫ e^u/u du - [-uˉ¹e^u + ∫uˉ¹e^u du]) =

1/(2e)(∫ e^u/u du + e^u/u - ∫e^u/u du) =

Notice that the integrals disappear! And you're left with:

e^u/2ue + C =

e^(u - 1)/2u + C =

e^(x² + 1 - 1)/[2(x² + 1)] + C =

e^(x²)/[2(x² + 1)] + C

= = = = = = = = = = = = = = = = = = = = =

Double check by differentiating:

[2(x² + 1)e^(x²)(2x) - e^(x²)(4x)]/4(x² + 1)² =

4x[(x² + 1)e^(x²) - e^(x²)]/4(x² + 1)² =

x[(x²e^(x²) + e^(x²) - e^(x²)]/(x² + 1)² =

x³e^(x²)/(x² + 1)²

It worked!

Answer 2

that integral can't be integrated, at least not indefinitely. Nothing on integration tables like it at all, because of the fact that the x^3 multiplies e^(x^2) makes it so much more complex. I've tried mathematica, TI-89, TI-Derive--- and nope, nothing

Answer 3

Damn this one is tuff. (for me) I'm working on it right know. I did some similar problems and forgot how to do them. I don't see a immediated u substituiton and other integration forms are still froeign to me. I can't see how ln can work yet, maybe there's a simple way of doing it that just looks hard.

Answer 4

(x^3 * e^x)' = 3x^2 * e^x + x^3 * e^x
[(x^2 + 1)^2 ] ' = 4x(x^2 +1)

thus :

[(x^3 * e^x) /(x^2 + 1)^2 ] '
=[ e^x*x^2 (3 + x)(x^2 + 1)^2 - 4x(x^2 +1)*(x^3 * e^x) ] / (x^2 + 1)^4

Answer 5

dang, I'm in cal 3 and I can't u-substitue, can't get anywhere with partial fraction decomp...I have no idea...

also couldn't get anywhere with integration by parts

Answer 6

You post it on Yahoo Answers!!