How to find the integral of x^3 e^(x^2) / (x^2 + 1)^2 ?

[rocomath]

Ok wow, I can't solve this :(

\int\frac{x^3e^{x^2}}{(x^2+1)^2}dx

I have let u equal almost everything possible.

u=x^2
d=2xdx

u=e^{x^2}
\ln u=x^2
\frac{du}{u}=2xdx

u=x^2+1 \leftrightarrow u-1=x^2
\ln u=\ln{(x^2+1)}
\frac{du}{u}=\frac{2xdx}{x^2+1}

 

[Acetv]

rocomath, is the integral given as a definite integral? It is possible to find the value of the definite integral by using integration by parts (think about the change of variables a bit). I don't really see an easy u-substitution that's going to help you out, but maybe I'm just blind.

 

[rocomath]

OH, lol. I forgot to include what dV was ...

But yes, I've been trying parts on it for a while now ... still nothing!

 

[tiny-tim]

Hi rocomath!

Go with u=x^2 again …

you should get \int\frac{ue^u}{(1 + u)^2}du

Then use partial fractions, and try the obvious.

 

[rocomath]

Hi rocomath!

Go with u=x^2 again …

you should get \int\frac{ue^u}{(1 + u)^2}du

Then use partial fractions, and try the obvious.

Uh! I didn't even think of using Partial Fractions bc the people in the group said they weren't doing that yet ... so frustrated, lol.

tiny-tim you're a life saver!

 

[rocomath]

Ok ... so I have

\frac 1 2\int\frac{te^t}{(t+1)^2}dt

Now using Partial Fractions and leaving the constant 1/2 out for now:

\frac{te^t}{(t+1)^2}=\frac{A}{t+1}+\frac{B}{(t+1)^2}

te^t=A(t+1)+B

if t = -1

B=-\frac 1 e

I'm having a hard time solving for A ...

I know what B is, so if t = 0

A=-\frac 1 e ?

... hmm, definitely not right!

I've never done a problem that had a product on the left side.

 

[Acetv]

The method of partial fractions won't work like it normally does here since you can't find integers A and B such that

t e^t = A (t + 1) + B

You can, however, solve for A and B in terms of your variable t, but it doesn't really make the problem any easier. We might find

B = - e^{-1}

And so plugging that into the equation above we find

A = \frac{t e^t - e^{-1}}{t + 1}

I still suggest trying the method of integration by parts. Hint: can you get the entire denominator (x^2+1)^2 into dv such that you can easily integrate it?

 

[rocomath]

Ok, got it!

Uh, sucha killer. Thanks for the hint Acetv, whewww!

 

[Gib Z]

We still could have used ordinary partial fractions - I'm not sure why you thought it was even possible to find constants A and B for te^t = A(t+1) + B, but tiny-tims suggestion was to use partial fractions on what you've been taught to use it on - Rational functions.

IE Only the \frac{t}{(t+1)^2}. After that, multiply each term by e^t.

You should have got \frac{e^t}{t+1} - \frac{e^{t}}{(t+1)^2} = \frac{1}{e} \left( \frac{e^u}{u} - \frac{e^u}{u^2} \right) where u = t+1, which are simple to integrate.