Question

Prove that $\frac{dx}{x^2-a^2} =\frac{1}{2a} log |\frac{x-a}{x+a}| +c$

Answer 1

$\bf \color{brown}{ \underline{ \underline{Step \: \: by \: \: step \: \: explanation : }}} \\ \\ \tt \int \frac{1}{ {x}^{2} - {a}^{2} } dx \\ \\ \\ \implies \tt\int \frac{1}{(x - a)(x + a)} \\ \\ \implies \tt \int \{\frac{1}{2a} ( \frac{1}{x - a} - \frac{1}{x + a} ) \}\\ \\ \implies \tt \frac{1}{2a} ( \int \frac{dx}{x - a} - \int \frac{dx}{x + a} ) \\ \\ \implies \tt \frac{1}{2a} \{ log(x - a) - log(x + a) \} + c \\ \\ \implies { \green {\boxed{\tt \frac{1}{2a} log( \frac{x - a}{x + a} ) + c}}} \bf \: \: \: \: \: proved \\ \\ \\ \\ \LARGE \mathfrak \red{Merry \: \: Christmas} \\ \\ \\ \\ \\ \tt \colorbox{aqua}{@StayHigh}$

Answer 2

Answer:

please mark as brainlist

Step-by-step explanation:

here is the answer