int frac{dx}{a^2 - x^2} is equal to | vedclass

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(C) To evaluate the integral $I = \int \frac{dx}{a^2 - x^2}$,we use partial fraction decomposition.We can write the integrand as:$\frac{1}{a^2 - x^2}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) To evaluate the integral $I = \int \frac{dx}{a^2 - x^2}$,we use partial fraction decomposition.We can write the integrand as:$\frac{1}{a^2 - x^2} = \frac{1}{(a - x)(a + x)} = \frac{1}{2a} \left( \frac{1}{a + x} + \frac{1}{a - x} \right)$.Now,integrating both sides with respect to $x$:$I = \frac{1}{2a} \int \left( \frac{1}{a + x} + \frac{1}{a - x} \right) dx$$I = \frac{1}{2a} \left( \int \frac{1}{a + x} dx + \int \frac{1}{a - x} dx \right)$$I = \frac{1}{2a} \left( \log |a + x| - \log |a - x| \right) + C$Using the logarithmic property $\log m - \log n = \log \left( \frac{m}{n} \right)$:$I = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C$.Thus,the correct option is $C$.

$\int \frac{dx}{a^2 - x^2}$ is equal to

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