Solving Integrals Using Substitution

[Lancelot59]

I'm attempting to solve the following problem:

[tex]\int_{0}^{\infty} {\frac{x arctan(x)}{(1+x^{2})^{2}}dx}[/tex]

I started with a substitution:

[tex]u=arctan(x), du=\frac{1}{(1+x^{2})}dx[/tex]

This seemed like the right thing to do, but after trying to put it together in several different ways I got nowhere. I looked at what WolframAlpha had to say. It got this after doing the same substitution:

[tex]\int_{}^{} {u sin(u)cos(u)du}[/tex]

I've gone at this for over half an hour now and I've gotten nowhere. Some insight into how this step was made would be appreciated.

 

[w3390]

What will happen if you make u=(1+x^2)^2?

 

[Lancelot59]

I got this:

[tex]\int_{}^{} {\frac{arctan(x)}{u*4\sqrt{u}du}[/tex]
which can be:
[tex]\int_{}^{} {\frac{arctan(x)}{4u^{\frac{3}{2}}}du}[/tex]

It's not incredibly helpful. I see no way of dealing with that arctangent.

 

[Tedjn]

If u = arctan(x), what does x equal?

 

[Mark44]

I'm attempting to solve the following problem:

[tex]\int_{0}^{\infty} {\frac{x arctan(x)}{(1+x^{2})^{2}}dx}[/tex]

I started with a substitution:

[tex]u=arctan(x), du=\frac{1}{(1+x^{2})}dx[/tex]

There is still a factor of x in the numerator that is unaccounted for, so this substitution isn't feasible.

This seemed like the right thing to do, but after trying to put it together in several different ways I got nowhere. I looked at what WolframAlpha had to say. It got this after doing the same substitution:

[tex]\int_{}^{} {u sin(u)cos(u)du}[/tex]

I've gone at this for over half an hour now and I've gotten nowhere. Some insight into how this step was made would be appreciated.

Use integration by parts, with u = arctan(x) and dv = xdx/((1 + x^2)^2). The resulting integral can be evaluated using a trig substitution.

The original integral is improper because of the infinity as one of the limits of integration, so you will need to take a limit at some point. One way to go about this is to evaluate this integral:
[tex]\int_0^b \frac{x~arctan(x)~dx}{(1 + x^2)^2}[/tex]

Your result of this integral will involve b, so take the limit as b goes to infinity to get your final answer.

 

[Lancelot59]

I'll try that solution out. I usually solve the indefinite integral first, then work out the proper solution for the definite integral with the limit.
[tex]-\frac{arctan(x)}{2(1+x^{2})}+\frac{1}{2}\int_{}^{} {(\frac{1}{(1+x^{2})})^{2}dx}[/tex]

I don't see a trig substitution working for that integral. I'll keep working at it.

 

[rl.bhat]

I'll try that solution out. I usually solve the indefinite integral first, then work out the proper solution for the definite integral with the limit.
[tex]-\frac{arctan(x)}{2(1+x^{2})}+\frac{1}{2}\int_{}^{} {(\frac{1}{(1+x^{2})})^{2}dx}[/tex]

I don't see a trig substitution working for that integral. I'll keep working at it.

[tex]\int_{}^{}(\frac{1}{(1+x^{2})^2}dx[/tex]

Substitute x = tanθ, the dx = sec^2(θ)dθ

Integration becomes

[tex]\int_{}^{}cos^2(\theta)d(\theta)[/tex]

Substitute [tex]cos^2(\theta)= \frac{1}{2}(1 + cos2(\theta))[/tex]

Now solve the integration.

 

[Lancelot59]

Yup, I managed to get it. Thanks for the help.