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Integration by Substitution

Solve: ∫1a2 ...

Question

Solve:
$\int \frac{1}{a^{2} - x^{2}} = \frac{1}{2 a} log ∣ ∣ ∣ \frac{a + x}{a - x} ∣ ∣ ∣$

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Solution

We have,

$I = \int \frac{d x}{a^{2} - x^{2}}$

$I = - \int \frac{d x}{(x - a) (x + a)}$

$I = - \frac{1}{2 a} [\int \frac{d x}{(x - a)} - \int \frac{d x}{(x + a)}]$

$I = - \frac{1}{2 a} [ln (x - a) - ln (x + a)] + C$

$I = \frac{1}{2 a} [ln | x + a | - ln | x - a |] + C$

$I = \frac{1}{2 a} [ln | a + x | - ln | a - x |] + C$

$I = \frac{1}{2 a} [ln ∣ ∣ ∣ \frac{a + x}{a - x} ∣ ∣ ∣] + C$

Hence, proved.

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\frac{d}{d x} [\frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} {sin}^{- 1} \frac{x}{a}] =

Q. How many of the following integrals are correct?
1.

\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = l n | x + \sqrt{x^{2} + a^{2}} | + C

\int \frac{d x}{\sqrt{x^{2} - a^{2}}} = l n | x - \sqrt{x^{2} - a^{2}} | + C

\int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} l n ∣ ∣ \frac{x - a}{x + a} ∣ ∣ + C

\int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} l n ∣ ∣ \frac{x + a}{x - a} ∣ ∣ + C

log (\frac{x + \sqrt{x^{2} - a^{2}}}{x - \sqrt{x^{2} - a^{2}}})

x : | x + 1 | = | x | + a, a \neq \pm 1

\frac{x + \sqrt{x^{2} - 1}}{x - \sqrt{x^{2} - 1}} + \frac{x - \sqrt{x^{2} - 1}}{x + \sqrt{x^{2} - 1}} = 98.

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