Question #5c331

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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#