`int(xtan^-1x)/(1+x^2)^(3/2)dx`

1.	`int(xtan^-1x)/(1+x^2)^(3/2)dx`
Answer» Put x=tan t so that dx `=sec^(2)tdt`. `:.int(xtan^(-1)x)/((1+x^(2))^(3//2))dx=int((tant)t)/((1+tan^(2)t)^(3//2))sec^(2)tdt` `int((tant)t)/(sect)dt=inttsintdt` `=t(-cost)-int1(-cost)-int1*(1-cost)dt" "` [integrating by parts] `=-tcost+sint+C=(-tan^(-1)x)/(sqrt(1+x^(2)))+(x)/(sqrt(1+x^(2)))+(x)/(sqrt(1+x^(2)))+C` `[becausesint=(x)/(sqrt(1+x^(2)))andcost=(1)/(sqrt(1+x^(2)))]`.

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