MHB Integral: Wolfram|Alpha - Solving Steps Explained

[leprofece]

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

 

[Opalg]

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.)