Integral: Wolfram|Alpha - Solving Steps Explained

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SUMMARY

The discussion focuses on the integral of the function \(\frac{x \arctan x}{(1+x^2)^3}\) and the steps involved in solving it using Wolfram|Alpha. Participants clarify the substitution of \(u = \arctan x\), leading to the transformation of the integral into \(\int u \sin u \cos^3 u \, du\). The importance of consistently substituting all instances of \(x\) with \(u\) is emphasized to avoid confusion in the integration process.

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ntegral (x.arctgx)/(1+x^2)^3?
int - Wolfram|Alpha...

i don't understand where int u sin u cos^3 (u) come from?

I HAVE JUST ASKED becauise i don't understand the step by step of the wilfrang alpha

ok change the variagle u = arctg (x) derive substitute and get to (u x)/(1+x^2)^2 trigonemtetry in nowhwere
 
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leprofece said:
ntegral (x.arctgx)/(1+x^2)^3?
int - Wolfram|Alpha...

i don't understand where int u sin u cos^3 (u) come from?

I HAVE JUST ASKED becauise i don't understand the step by step of the wilfrang alpha

ok change the variable u = arctg (x) derive substitute and get to (u x)/(1+x^2)^2 trigonemtetry in nowhwere
If $u = \arctan x$ then $x = \tan u$, and $$\int \frac{x\arctan x}{(1+x^2)^3}dx$$ becomes $$\int \frac{u\tan u}{(1 + \tan^2u)^3}\sec^2u\,du$$, which simplifies to $$\int u \sin u \cos^3u\,du.$$ (When you make the substitution, you must re-write all the $x$'s in terms of $u$, not just some of them.)
 

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