Question

Evaluate :-
$\displaystyle \sf\int \dfrac{1}{ {x}^{3} - 1}dx$
​

Answer 1

EXPLANATION.

$\implies \displaystyle \int \dfrac{dx}{x^{3} - 1}$

As we know that,

We can write equation as,

⇒ x³ - 1 = x³ - 1³.

Apply the formula of :

⇒ (a³ - b³) = (a - b)(a² + ab + b²).

⇒ (x³ - 1³) = (x - 1)(x² + x + 1).

$\implies \displaystyle \int \dfrac{dx}{(x - 1)(x^{2} + x + 1)}$

$\implies \dfrac{1}{(x - 1)(x^{2} + x + 1)} \ = \dfrac{A}{(x - 1)} \ + \dfrac{Bx + C}{(x^{2} + x + 1)}$

$\implies 1 =A(x^{2} + x + 1) \ + (Bx + C)(x - 1)$

$\implies 1 = Ax^{2} + Ax + A + Bx^{2} - Bx + Cx - C.$

$\implies 1 = Ax^{2} + Bx^{2} + Ax - Bx + Cx + A - C.$

$\implies 1 = (A + B)x^{2} + (A - B + C)x + (A - C)$

Compare the coefficient of the equation, we get.

$\implies A + B = 0. - - - - - (1).$

$\implies A - B + C = 0. - - - - - (2).$

$\implies A - C = 1. - - - - - (3).$

$\implies B = A + C . - - - - - (4).$

$\implies B = -A. - - - - - (5).$

$\implies A = 1 + C. - - - - - (6).$

Put the value of equation (4) in equation (5), we get.

$\implies A + C = -A.$

$\implies C = -2A. - - - - - (7).$

Put the value of equation (7) in equation (6), we get.

$\implies A = 1 - 2A.$

$\implies 3A = 1.$

$\implies A = \dfrac{1}{3}$

Put the value of A = 1/3 in equation (5), we get.

$\implies B = -\dfrac{1}{3}$

Put the value of A = 1/3 in equation (7), we get.

$\implies C = -\dfrac{2}{3}$

$\implies \displaystyle \int \dfrac{A}{(x - 1)} dx \ + \int \dfrac{Bx + C}{x^{2} + x + 1}$

$\implies \displaystyle \int \dfrac{dx}{3(x - 1)} \ + \int \dfrac{\bigg(\dfrac{-1}{3} x\bigg)+ \bigg(\dfrac{-2}{3} \bigg)}{x^{2} + x + 1} dx$

$\implies \displaystyle \dfrac{1}{3} ln|x - 1| \ - \dfrac{1}{3} \int \dfrac{x + 2}{x^{2} + x + 1} dx$

Multiply and divide second equation by 2, we get.

$\implies \displaystyle \dfrac{1}{3} ln|x - 1| \ - \dfrac{1}{6} \int \dfrac{2x + 4}{x^{2} + x + 1} dx$

$\implies \displaystyle \dfrac{1}{3} ln|x - 1| \ - \dfrac{1}{6} \int \dfrac{2x + 1 + 3}{x^{2} + x + 1} dx$

$\implies \displaystyle \dfrac{1}{3} ln|x - 1| \ - \dfrac{1}{6} \int \dfrac{2x + 1}{x^{2} + x + 1} dx\ - \dfrac{3}{6} \int \dfrac{dx}{x^{2} + x + 1}$

In second equation,

Using substitution in equation, we get.

⇒ x² + x + 1 = t.

Differentiate w.r.t x, we get.

⇒ 2x + 1 dx = dt.

$\implies \displaystyle \dfrac{1}{3} ln|x - 1| \ - \dfrac{1}{6} \int \dfrac{dt}{t} \ - \dfrac{1}{2} \int \dfrac{dx}{x^{2} + x + 1}$

Put the value of t = x² + x + 1 in the equation, we get.

$\implies \displaystyle \dfrac{1}{3} ln|x - 1| \ - \dfrac{1}{6} ln |x^{2} + x + 1| \ - \dfrac{1}{2} \int \dfrac{dx}{x^{2} + x + 1}$

From equation 3rd we can apply perfect square method, we get.

⇒ x² + x + 1 = (x + 1/2)² + (√3/2)².

Using this formula in the equation, we get.

$\implies \displaystyle \dfrac{1}{3} ln|x - 1| \ - \dfrac{1}{6} ln |x^{2} + x + 1| \ - \dfrac{1}{2} \int \dfrac{dx}{\bigg(x + \dfrac{1}{2} \bigg)^{2} + \bigg(\dfrac{\sqrt{3} }{2} \bigg)^{2} }$

As we know that,

Formula of :

$\implies \displaystyle \int \dfrac{dx}{x^{2} + a^{2} } \ = \dfrac{1}{a} tan^{-1} \bigg(\dfrac{x}{a} \bigg) + C.$

Using this formula in the 3rd equation, we get.

$\implies \displaystyle \dfrac{1}{3} ln|x - 1| \ - \dfrac{1}{6} ln |x^{2} + x + 1| \ - \dfrac{1}{2} \times \dfrac{1}{\bigg(\dfrac{\sqrt{3} }{2} \bigg)} tan^{-1} \bigg(\dfrac{(x + 1/2)}{(\sqrt{3} /2)} \bigg) + C.$

$\implies \displaystyle \dfrac{1}{3} ln|x - 1| \ - \dfrac{1}{6} ln |x^{2} + x + 1| \ - \dfrac{1}{\sqrt{3} } tan^{-1} \bigg(\dfrac{2x + 1}{\sqrt{3} } \bigg) + C.$

Answer 2

Step-by-step explanation:

this is the ANSWER

amanshrama264 is correct