Integrate (x^2)/(1+x^2)^2

[RiskaY]

Hi everyone,

I would greatly appreciate if someone could point me in the right direction about how to evaluate: Integral: (x^2)/(1+x^2)^2

I found the solution in the table of integrals, but this is unacceptable in homework. Any help is appreciated!!!

 

[galactus]

There are different approaches, but you could use partial fractions.

\(\displaystyle \L\\\frac{x^{2}}{(1+x^{2})^{2}}=\frac{1}{1+x^{2}}-\frac{1}{(1+x^{2})^{2}}\)

\(\displaystyle \L\\\int\frac{1}{1+x^{2}}dx-\int\frac{1}{(1+x^{2})^{2}}dx\)

You should recognize the first one as being arctan. For the second part, try letting \(\displaystyle x=tan(u),\;\ dx=sec^{2}(u)du\). Can you proceed?.

 

[soroban]

Hello, RiskaY!

I found two more ways . . .

\(\displaystyle \L \int\frac{x^2}{(1\,+\,x^2)^2}\,dx\)

Trig Substitution

Let \(\displaystyle x\,=\,\tan\theta\;\;\Rightarrow\;\;dx\,=\,\sec^2\theta\,d\theta\)

Substitute: \(\displaystyle \L\:\int\frac{\tan^2\theta}{(\sec^2\theta)^2}\,(\sec^2\theta\,d\theta)\;=\;\L\int\frac{\tan^2\theta}{\sec^2\theta}\,d\theta \;=\;\L\int\)\(\displaystyle \sin^2\theta\,d\theta\)

We have: \(\displaystyle \:\frac{1}{2}\L\int\)\(\displaystyle (1\,-\,\cos2\theta)\,d\theta \;= \;\frac{1}{2}\left(\theta\,-\,\frac{1}{2}\sin2\theta\right)\,+\,C \;= \;\frac{1}{2}\left(\theta\,-\,\sin\theta\cos\theta\right)\,+\,C\)

Back-substitute: \(\displaystyle \L\:\frac{1}{2}\left(\arctan x\,-\,\frac{x}{1\,+\,x^2}\right) \,+\,C\)

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By Parts

\(\displaystyle \text{Let: }\,u\,=\,x\;\;\;\;\;dv\:=\:\frac{x}{(1\,+\,x^2)^2}\:=\:x(1\,+\,x^2)^{-2}dx\)

\(\displaystyle \text{Then: }\,du\,=\,dx\;\;\;v\:=\:-\frac{1}{2}(1\,+\,x^2)^{-1}\:=\:-\frac{1}{2}\,\frac{1}{1\,+\,x^2}\)


We have: \(\displaystyle \L\:-\frac{1}{2}\,\frac{x}{1\,+\,x^2} \,+\,\frac{1}{2}\int\frac{dx}{1\,+\,x^2}\)

. . . . . . \(\displaystyle \L=\;-\frac{1}{2}\,\frac{x}{1\,+\,x^2}\,+\,\frac{1}{2}\,\arctan x\,+\,C\)

. . . . . . \(\displaystyle \L=\;\frac{1}{2}\left(\arctan x\, -\, \frac{x}{1\,+\,x^2}\right)\,+\,C\;\;\) . . . ta-DAA!

 

[RiskaY]

Thanks a million guys!