Question

integration of (x³)*(tan-¹ x)dx

Answer 1

here is your answer

 

Let       u = tan-1x                      dv = x3dx

         du = 1 / (1 + x2)dx             v = (1/4)x4

 

Then the integral is

 

(1/4)x4tan-1x - (1/4)∫[x4 / (1 + x2)]dx

 

 

Then, you take the integral of the sub-integral.

 

∫x4 / (1 + x2)dx

 

using trig substitution.

 

x = tanθ                        x2 = tan2θ

dx = sec2θdθ                 x4 = tan4θ

 

∫[(tan4θ) / (1 + tan2θ)] sec2θdθ

 

Since              sec2θ = 1 + tan2θ

 

∫tan4θdθ