Integrate \;{1}{x²-a²} is

Correct Answer is : 12alog⁡∣ x−ax+a∣+k\;1/2a ≤ft| \;x-a/x+a |+k2a1logx+ax−a+k

Correct Answer is : 12alog⁡∣ x−ax+a∣+k\;1/2a ≤ft| \;x-a/x+a |+k2a1logx+ax−a+k

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Integral Calculus

Question : 55 of 73

Marks: +1, -0

Integrate

\;\frac{1}{x^{2}-a^{2}}

Solution:

Concept:

The integral of

\frac{1}{x^2 - a^2}

involves partial fractions and a standard formula.

Explanation:

We express

\frac{1}{x^2 - a^2}

\frac{1}{(x-a)(x+a)}

Using partial fractions,

\frac{1}{(x-a)(x+a)} = \frac{1}{2a}\left(\frac{1}{x-a} - \frac{1}{x+a}\right)

Integrating,

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

Thus,

\int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right| + k

Answer:

\frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + k

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