INT 1 x3 1 dxSolutionI 1 x 1 dx Let I 1 x1xx1

Solution

I = [ 1 / ( x³ - 1 ) ] dx

Let :

I = { 1 / (x-1)(x²+x+1) ] } dx = { [ A / (x-1) ] + [ (Bx+C) / (x²+x+1) ] } dx ... (1)

Then,

... A(x²+x+1) + (Bx+C)(x-1) 1.......... (2)

x = 1 A(3) + 0 = 1 A = 1/3

Also, differentiating (2) w.r.t. x,

A(2x+1) + (Bx+C)(1) + B(x-1) = 0 ......... (3)

x = 0 A(1) + C - B = 0 B - C = 1/3 ..... (4)

Differentiating (3) w.r.t. x,

A(2) + B(1) + B(1) = 0 B = -A = -1/3

from (3), C = B - (1/3) = (-1/3) - (1/3) = -2/3
......................................…

From (1), then,

I = (1/3) [ 1 / (x-1) ] dx + { [ (-1/3)x - (2/3) ] / (x²+x+1) } dx

= ....... the main job is done ........

= ....... now you do the rest ..........