Question #5c331

Question

1236 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-01T18:26:44

$=1/2( x^2 tan^(-1) x + tan^(-1) x - x)+ C$

$I = int \ x tan^(-1) x \ dx$

we'll use IBP

before we go, note that $d/dx (tan^(-1) x) = 1/(1+x^2)$ , a well known result

So
$I = int \ d/dx(x^2/2) tan^(-1) x \ dx$

and by IBP
$I = x^2/2 tan^(-1) x - int \ x^2/2 d/dx( tan^(-1) x) \ dx$

$= x^2/2 tan^(-1) x - 1/2 int \ x^2 1/(x^2 +1 ) \ dx$

$= x^2/2 tan^(-1) x - 1/2 int \ (x^2+ 1 - 1)/(x^2 +1 ) \ dx$

$= x^2/2 tan^(-1) x - 1/2 int \ 1 - 1/(x^2 +1 ) \ dx$

$= x^2/2 tan^(-1) x - 1/2 int \ 1 - d/dx (tan^(-1) x) \ dx$

$= x^2/2 tan^(-1) x - 1/2x + 1/2 tan^(-1) x + C$

$=1/2( x^2 tan^(-1) x + tan^(-1) x - x)+ C$