Question

Find integration:
∫ dx/x²-a²

Answer 1

Answer:

1/2a log |(x-a) /(x+a) |

Answer 2

$\int {\dfrac{1}{x^2-a^2} } \, dx$ $=\dfrac{1}{2a} \log [\dfrac{x-a}{x+a}] +C$

Step-by-step explanation:

We have,

$\int {\dfrac{1}{x^2-a^2} } \, dx$

To find, the value of $\int {\dfrac{1}{x^2-a^2} } \, dx$ = ?

∴ $\int {\dfrac{1}{x^2-a^2} } \, dx$

$=\int {\dfrac{1}{(x+a)(x-a)} } \, dx$

Using the identity,

$a^{2} -b^{2} =(a+b)(a-b)$

Multiplying numerator and denominator by 2a, we get

$=\dfrac{1}{2a} \int {\dfrac{2a}{(x+a)(x-a)} } \, dx$

$=\dfrac{1}{2a} \int {\dfrac{(x+a)-(x-a)}{(x+a)(x-a)} } \, dx$

$=\dfrac{1}{2a} [\int {\dfrac{(x+a)}{(x+a)(x-a)} } \, dx-\int {\dfrac{(x-a)}{(x+a)(x-a)} } \, dx]$

$=\dfrac{1}{2a} [\int {\dfrac{1}{(x-a)} } \, dx-\int {\dfrac{1}{(x+a)} } \, dx]$

$=\dfrac{1}{2a} [\log (x-a)-\log (x+a)]+C$

Where, C is called integration constant

Using the logarithm identity,

$\log \dfrac{a}{b} =\log a-\log b$

$=\dfrac{1}{2a} [\log (x-a)-\log (x+a)]+C$

$=\dfrac{1}{2a} \log [\dfrac{x-a}{x+a}] +C$

∴ $\int {\dfrac{1}{x^2-a^2} } \, dx$ $=\dfrac{1}{2a} \log [\dfrac{x-a}{x+a}] +C$