Question

Integral is a pure fun (HELP)!

Answer 1

\[\int\limits \frac{ x^2 }{ (1+x^2)^2 }dx\]

Answer 2

@rvc

Answer 3

any hint?

Answer 4

I would guess it has something to do woth arctan

Answer 5

with*

Answer 6

@IrishBoy123 ?

Answer 7

\[\large \int \frac{x^2 \color{red}{+1-1}}{(x^2+1)^2}dx\] -My first guess.

Second guess would be partial fractions maybe?

Answer 8

I've tried that

Answer 9

same @Snuffleupagas i thought
well i think it would be more easy if at all it had e^x

Answer 10

x = tan theta and then a trig switch via double angle formula

Answer 11

\[x=\tan(\theta)~,~ dx = \sec^2(\theta)d\theta\ \]\[\int \frac{\tan^2(\theta)}{(\tan^2(\theta)+1)^2}\cdot \sec^2(\theta)d\theta\ \]\[\color{red}{\tan^2(\theta)+1=\sec^2(\theta)}\]\[\int \frac{\tan^2(\theta)}{(\sec(\theta))^4} \cdot \sec^2(\theta) d\theta\ \]

Answer 12

I see how @IrishBoy123 spotted the trig sub, he saw \[\int \left(\color{blue}{\frac{x}{x^2+1}}\right)^2dx\] which relates to \(\color{blue}{\tan^{-1}(\theta)}\)

Answer 13

idk guys I don't see a way out here

Answer 14

hmmm

Answer 15

Don't give up! Work off of whats here :)

\(\color{blue}{\text{Originally Posted by}}\) @Snuffleupagas
\[x=\tan(\theta)~,~ dx = \sec^2(\theta)d\theta\ \]\[\int \frac{\tan^2(\theta)}{(\tan^2(\theta)+1)^2}\cdot \sec^2(\theta)d\theta\ \]\[\color{red}{\tan^2(\theta)+1=\sec^2(\theta)}\]\[\int \frac{\tan^2(\theta)}{(\sec(\theta))^4} \cdot \sec^2(\theta) d\theta\ \]
\(\color{blue}{\text{End of Quote}}\)

Answer 16

yep
u will get \[\int\limits \sin^2\theta\ d \theta\]

Answer 17

Remember, \(\tan^2(\theta) = \sec^2(\theta) - 1\)

Answer 18

\[\large \int \frac{\sec^2(\theta)-1}{\sec^2(\theta)}d\theta\]\[\large \int \left[\frac{\sec^2(\theta)}{\sec^2(\theta)} -\frac{1}{\sec^2(\theta)}\right]d\theta\]

Answer 19

I think I got that!

Answer 20

Congrats !!

Answer 21

Great :D Give that medal to @IrishBoy123 for dropping by a care package :)

Answer 22

Thank you all for your patience !

Answer 23

you've opened my eyes!