The value of \; {dx}{x(x²-1)} is equal to

Correct Answer is : 12log⁡(1− 1x2)+k1/2 ≤ft(1-\;{1}{x²})+k21log(1−x21)+k

Correct Answer is : 12log⁡(1− 1x2)+k1/2 ≤ft(1-\;{1}{x²})+k21log(1−x21)+k

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Integral Calculus

Question : 42 of 73

Marks: +1, -0

The value of

\int\; \frac{dx}{x(x^{2}-1)}

is equal to

Solution:

Concept:

Integrate using partial fractions and simplify to match the given form.

Explanation:

Rewrite the integrand:

\frac{1}{x(x^{2}-1)} = \frac{1}{x(x-1)(x+1)}

Use partial fractions:

\frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}

Solving gives

A = -1

B = \frac{1}{2}

C = \frac{1}{2}

Integrate:

\int \frac{dx}{x(x^{2}-1)} = -\int \frac{dx}{x} + \frac{1}{2}\int \frac{dx}{x-1} + \frac{1}{2}\int \frac{dx}{x+1}

= -\ln|x| + \frac{1}{2}\ln|x-1| + \frac{1}{2}\ln|x+1| + k

Combine logs:

\frac{1}{2}\ln\left|\frac{(x-1)(x+1)}{x^2}\right| = \frac{1}{2}\ln\left|1 - \frac{1}{x^2}\right| + k

Thus

\int \frac{dx}{x(x^{2}-1)} = \frac{1}{2}\ln\left(1 - \frac{1}{x^{2}}\right) + k

Answer:

Option B:

\frac{1}{2} \log \left(1-\;\frac{1}{x^{2}}\right)+k

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