Finding $\int x\arctan(x)\,dx$ through inverse substitution

I have to find $\int x\arctan(x)\,dx$ through inverse substitution.

My attempt:

Let $x = \tan\theta$, and $dx = \sec^2\theta \,d\theta$. So, we have that

$\int x\arctan(x)\,dx = \int \tan(\theta)\arctan(\tan(\theta))sec^2\theta d\theta = \int \tan(\theta)\theta \sec^2(\theta) = \int \theta \tan(\theta)(\tan^2(\theta) + 1)d\theta$.

That's about as far as I could get. I tried substituting or integrating by parts after I got to the last step, but nothing really worked.

EDIT: The problem states that we must integrate by inverse substitution.

Thanks.

• Surely parts works. Just try the less intuitive direction...integrate $x$ and differentiate $\arctan x$. That gives you something like $\int \frac {x^2}{1+x^2} dx$. No problem.
– lulu
Feb 25 '18 at 4:07
• But I have to use inverse substitution as its a requirement of the problem. I tried doing it by parts after I got to my last step, and nothing really worked as far as I can see.
– user462562
Feb 25 '18 at 4:08
• Well, my way has an inverse substitution at some point! I mean, you just write $\frac {x^2}{1+x^2}=1-\frac {1}{1+x^2}$ and I need an inverse substitution to do the last bit! (ok, I really don't but artificial problems call for artificial solutions).
– lulu
Feb 25 '18 at 4:11
• I haven't used theta substitution ,so I hope that won't be a problem for your question.
– user517784
Feb 25 '18 at 4:35

If you must integrate by inverse substitution, just as you did using $x=\tan (t)$, $dx=\sec ^2(t)\,dt$ you effectively end with $$\int x \tan ^{-1}(x)\,dx=\int t \tan (t) \sec ^2(t)\,dt$$ Now, one integration by parts $$u=t \implies u'=dt$$ $$v'=\tan (t) \sec ^2(t)\,dt\implies v=\frac{1}{2}\sec ^2(t)$$ leading to $$\frac{t}{2}\sec ^2(t)-\frac{1}{2}\int\sec ^2(t)\,dt=\frac{t}{2}\sec ^2(t)-\frac{1}{2}\tan(t)$$

Take $u= tan^{-1}(x)$. Then$u'=1/(1+x^2)$.Let $v' =x ,v=x^2/2$.

Here we use the inverse substitution

Therefore on integrating we get:-

Solving using integration by parts.

$x^2 \frac {tan^{-1}(x)}{2} -\frac{1}{2} \int \frac{ x^2 dx}{(1+x^2)}$

Skipping the remaining steps,if you want I can add them too

$\frac{x^2 tan^{-1}(x)}{2} - \frac {x}{2} + \frac {tan^{-1}(x)}{2} +C$