Question

What is \(\rm \int \frac{dx}{x\left ( x^{3}-1 \right )}\) equal to ? 
1. \(\rm \log\left | \frac{x^{3}-1}{x^{3}} \right | + C\)
2. \(\rm \frac{1}{3}log\left | \frac{x^{3}-1}{x^{3}} \right | + C\)
3. \(\rm \frac{1}{3}log\left | \frac{x^{3}+1}{x^{3}} \right | + C\)
4. \(\rm \frac{1}{3}log\left | \frac{x^{3}-1}{x^{3} + 1} \right | + C\)

Answer 1

Correct Answer - Option 2 : \(\rm \frac{1}{3}log\left | \frac{x^{3}-1}{x^{3}} \right | + C\)

Concept: 

\(\rm \int \frac{1}{x^{2}-a^{2}}dx = \frac{1}{2a}log\left | \frac{x-a}{x+a} \right |\)

Calculation: 

I = \(\rm \int \frac{dx}{x\left ( x^{3}-1 \right )}\) 

Let , x3- 1 = t          

Differentiate both sides w.r.t x 

⇒ 3x2 dx = dt 

⇒  \(\rm dx = \frac{dt}{3x^{2}}\) 

Substitute above values in given integration , 

I = \(\rm \frac 1 3 \int \frac{dt}{x^3t}\)

I = \(\rm \frac{1}{3}\int \frac{dt}{t\left ( t+1 \right )}\)          [∵ x3 = t + 1]

I = \(\rm \frac{1}{3}\int \frac{dt}{t^{2}+t}\) 

By completing square method , 

I = \(\rm \frac{1}{3}\int \frac{dt}{t^{2}+t+\frac{1}{4}-\frac{1}{4}}\) 

I = \(\rm \frac{1}{3}\int \frac{dt}{\left ( t+\frac{1}{2} \right )^{2}- \left ( \frac{1}{2} \right )^{2}}\) 

I = \(\rm \frac{1}{3}log\left | \frac{t}{t+1} \right | + C\) 

From eq. (i) 

I = \(\rm \frac{1}{3}log\left | \frac{x^{3}-1}{x^{3}} \right | + C\) .

The correct option is 2.