What is the best method for integrating x(arctanx)^2?

[Ayesh]

Homework Statement

I have to integrate using the following integration methods:
1) u-sub
2) by parts
3) trig
4) partial fraction

Homework Equations

$$\int$$x(arctanx)²

The Attempt at a Solution

According to me, the method I should use is integration by parts.

When I tried it, the whole thing got worse...

u = arctanx²
du = 2arctanx/1+x²
dv = xdx
v = 1/2x²

$$\int$$x(arctanx)² = 1/2x²(arctanx)² - $$\int$$x²arctanx/1+x₂

 

[Kreizhn]

I would use integration by parts, though you're going to need to use it twice.

First use IBP to find calculate $\int \arctan(x) dx$. Once you have this, you can use IBP on the whole thing.

 

[tt2348]

try using a U substitution, say u=arctan(x) and $$du=\frac{1}{1+x^{2}}dx$$
then dx=(1+x$$^{2}$$)du , where x=tan(u). also plugging in for the x multiplied in front gets you $$\int u^{2}tan(u)(1+tan(u)^{2})du$$

 

[Kreizhn]

try using a U substitution, say u=arctan(x) and $$du=\frac{1}{1+x^{2}}dx$$
then dx=(1+x$$^{2}$$)du , where x=tan(u). also plugging in for the x multiplied in front gets you $$\int u^{2}tan(u)(1+tan(u)^{2})du$$

That is definitely more elegant, though I'm not sure that Ayesh will see the trick involved to make it an easy IBP.

 

[Ayesh]

Kreizhn why should I use $$\int$$arctanx first?

 

[Ayesh]

I get from where dx(1 + tan(u)²)du comes from, but not the u²tan(u).

 

[Kreizhn]

If you're going to keep going in the more elegant direction, I wouldn't worry about integrating arctan first. It is a method that works, but takes a lot more work.

As for your other question, consider that your integrand is $x(\arctan(x))^2$ and you make the substitution $u = \arctan(x)$ or alternatively, $x = \tan(u)$. Now when making the substitution, there are three things you are going to need to account for, namely, the x, the $\arctan^2(x)$ and the dx.

see if you can follow the following bit of arithmetic

$$\begin{align*}<br /> x \arctan^2(x) dx &= x u^2 dx & \text{ since } u=\arctan(x) \\<br /> &= tan(u) u^2 dx & \text{ since } x = \tan(u) \\<br /> &= u^2 \tan(u) (1+x^2) du & \text{ since } dx = (1+x^2) du \\<br /> &= u^2 \tan(u) (1+tan^2(u) )du & \text{ since } x = \tan(u)<br /> \end{align*}$$

The tricky part now is realizing that there is an implicit derivative in here that will make integration by parts simple.

 

[Ayesh]

Thank you!

Your explanations are very clear!

 

[Ayesh]

This is what I have done until now:

$$\int$$x(arctanx)^2 dx

1/2 x²arctanx - integral 1/2x²(1/1+x^2) dx

1/2x²arctanx - 1/2 integral tan²t/1+tan²t *sec²t

1/2x²arctanx - 1/2 integral tan²t

1/2x²arctanx - 1/2 integral (sec²t - 1) dt

1/2x²arctanx - 1/2 integral (tant - t) dt

... ?

I don't what to do after.
I know it has something to do with drawing a triangle, but I don't know what values to put around it.