Substitute:
x=tantx=tant
dx = (dt)/cos^2tdx=dtcos2t
intx^3/(1+x^2)dx = int tan^3t/(1+tan^2t) dt/cos^2t∫x31+x2dx=∫tan3t1+tan2tdtcos2t
Use the trigonometric identity:
1+tan^2 t = 1/cos^2t1+tan2t=1cos2t
intx^3/(1+x^2)dx = int tan^3t/( 1/cos^2t) dt/cos^2t = int tan^3tdt∫x31+x2dx=∫tan3t1cos2tdtcos2t=∫tan3tdt
Using the same identity again:
int tan^3tdt = int tant tan^2t dt = int tant (1/cos^2t -1)dt =∫tan3tdt=∫tanttan2tdt=∫tant(1cos2t−1)dt=
= int tant d(tant) - int tant dt =1/2 tan^2 t - int sint /cost dt ==∫tantd(tant)−∫tantdt=12tan2t−∫sintcostdt=
= 1/2tan^2t + int (d(cost))/cost = 1/2tan^2t + ln |cos t|=12tan2t+∫d(cost)cost=12tan2t+ln|cost|
To substitute back xx, we have that:
tant = xtant=x
cost= +-sqrt(1 /(1+tan^2t))= +-1/sqrt(1+x^2)cost=±√11+tan2t=±1√1+x2
So:
intx^3/(1+x^2)dx = 1/2x^2-lnsqrt(1+x^2)∫x31+x2dx=12x2−ln√1+x2
