Question

Integrate the following w.r.t. x : `(1)/(x(1 + 4x^3 + 3x^6)`

Answer 1

Let I = `int (1)/(x(1 + 4x^3 + 3x^6)).dx`

= `int x^2/(x^3(1 + 4x^3 + 3x^6)).dx`

Put x³ = t

∴ 3x² dx = dt

∴ `x^2dx = 1/3.dt`

∴ I = `1/3 int 1/(t(1 + 4t + 3t^2)).dt`

= `1/3 int 1/(t(t + 1)(3t + 1)).dt`

Let `1/(t(t + 1)(3t + 1)) = A/t + B/(t + 1) + C/(2t + 1)`

∴ 1 = A(t + 1)(3t + 1) + Bt(3t + 1) + Ct(t + 1)

Put t = 0, we get

1 = A(1) + B(0) + C(0)

∴ A = 1

Put t + 1 = 0, i.e. t = – 1 we get

1  = A(0) + B(– 1)(– 2) + C(0)

∴ B = `1/2`

Put 3t + 1 = 0,  i.e. t = `-1/3`, we get

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

∴ C = `-9/2`

∴ `1/(t(t + 1)(3t + 1)) = 1/t + ((1/2))/(t + 1) + ((-9/2))/(3t + 1)`

∴ I = `1/3 int[ 1/t + ((1/2))/(t + 1) + ((-9/2))/(3t + 1)].dt`

= `1/3[ int 1/t .dt + 1/2 int 1/(t + 1).dt - 9/2 int 1/(3t + 1).dt]`

= `1/3[log|t| + 1/2log|t + 1|- 9/2 . 1/3log|3t + 1|] + c`

= `1/3log|x^3| + 1/2 log|x^3 + 1| - 3/2 log|3x^3 + 1| + c`

= `log|x| + 1/2 log|x^3 + 1| - 3/2 log|3x^3 + 1| + c`.