# Indefinite Integration Challenge to the MMF Community

#### Etyucan

Here are some challenging problems which you may enjoy.

Problem 1

Evaluate $$\displaystyle \int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx$$

Problem 2

Evaluate $$\displaystyle \int\frac{2-x^2}{(1+x)\sqrt{1-x^2}}dx$$

Problem 3

Evaluate $$\displaystyle \int \frac{x^7-24x^4-4x^2+8x-8}{x^8+6x^6+12x^4+8x^2}dx$$

I can provide hints if required.

Have Fun!

#### agentredlum

Math Team
THIS POST IS OFF TOPIC

NICE AVATAR!!!!

-agent

#### mathbalarka

Math Team
The first one doesn't seems elementary to me, is it the case? The way I evaluated, I think it requires the elliptic integrals.

Balarka
.

#### Etyucan

Everything is elementary. The first one is merely

$$\displaystyle -2\arctan \sqrt{x+\frac{1}{x}+1}+C$$

And I don't see anything wrong with my post.

#### greg1313

Forum Staff
2. $$\displaystyle \int\frac{2\,-\,x^2}{(1\,+\,x)\sqrt{1\,-\,x^2}}\,dx$$

$$\displaystyle \sin(\theta)\,=\,x,\,\cos(\theta)\,d\theta\,=\,dx$$

$$\displaystyle \int\frac{$2\,-\,\sin^2(\theta)$\cos(\theta)}{$1\,+\,\sin(\theta)$\cos(\theta)}\,d\theta\,=\,\int\frac{1}{1\,+\sin(\theta)}\,+\,1\,-\,\sin(\theta)\,d\theta\,=\,\int\frac{1}{1\,+\,\sin(\theta)}\,d\theta\,+\,\theta\,+\,\cos(\theta)\,+\,C$$

$$\displaystyle \int\frac{1}{1\,+\,\sin(\theta)}\,d\theta\,\Leftrightarrow\,t\,=\,\tan\(\frac{\theta}{2}$$\,\Leftrightarrow\,\int\frac{2}{(1\,+\,t)^2}\,dt\,=\,-2(1\,+\,t)^{-1}\,+\,C\,=\,-2\frac{1}{1\,+\,\tan$$\frac{\theta}{2}$$}\,+\,C\)

$$\displaystyle -2\frac{1}{1\,+\,\tan\(\frac{\theta}{2}$$}\,+\,\theta\,+\,\cos(\theta)\,+\,C\,=\,-2\,\cdot\,\frac{1\,+\,\sqrt{1\,-\,x^2}}{1\,+\,x\,+\,\sqrt{1\,-\,x^2}}\,+\,\sin^{\small{-1}}(x)\,+\,\sqrt{1\,-\,x^2}\,+\,C\,=\,\int\frac{2\,-\,x^2}{(1\,+\,x)\sqrt{1\,-\,x^2}}\,dx\)

I used the Weierstrass Substitution.

#### CRGreathouse

Forum Staff
Etyucan said:
And I don't see anything wrong with my post.
If this is a response to agentredlum, I believe he was suggesting that *his* post was off-topic.

#### agentredlum

Math Team
CRGreathouse said:
Etyucan said:
And I don't see anything wrong with my post.
If this is a response to agentredlum, I believe he was suggesting that *his* post was off-topic.
Yes, MY post was off topic but i couldn't resist complimenting you about Seraph, from Matrix, as seen by NEO codevision, avatar.

Seraph protects that which matters most.

#### Etyucan

CRGreathouse said:
If this is a response to agentredlum, I believe he was suggesting that *his* post was off-topic.
OH! I am sorry. I thought he meant about my post.

#### galactus

Hey E. For the first one it is funny, if one runs it through Maple or Mathematica it returns this long convoluted Elliptic monstrosity.

I got something that is different than your arctan, but equivalent.

$$\displaystyle -\sin^{-1}\left(\frac{x^{2}+1}{(x+1)^{2}}\right)-\frac{\pi}{2}$$

Taking the derivative of this returns the integral you posted.

I rewrote it as $$\displaystyle \frac{1-x}{x(1+x)\sqrt{\frac{x^{2}+x+1}{x}}}$$

Made the sub $$\displaystyle x=\frac{1+t}{1-t}, \;\ dx=\frac{2}{(t-1)^{2}}dt$$

$$\displaystyle -2\int \frac{t}{\sqrt{(1-t^{2})(t^{2}+3)}}dt$$

$$\displaystyle -2\int\frac{t}{\sqrt{1-\left(\frac{t^{2}+1}{2}\right)^{2}}}dt$$

This is an arcsin integral which can be seen to evaluate to $$\displaystyle -\sin^{-1}\left(\frac{t^{2}+1}{2}\right)+C$$

Resubbing gives the final result along with $$\displaystyle C=\frac{-\pi}{2}$$

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