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Integration by Parts

Prove that ...

Question

Prove that $\int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} log | x + \sqrt{x^{2} - a^{2}} | + c$ .

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Solution

Given, $\int \sqrt{x^{2} - a^{2}} d x$
$= \sqrt{x^{2} - a^{2}} (x) - \int \frac{2 x}{2 \sqrt{x^{2} - a^{2}}} x d x$
$= \sqrt{x^{2} - a^{2}} (x) - \int \frac{x^{2}}{\sqrt{x^{2} - a^{2}}} d x$
$= \sqrt{x^{2} - a^{2}} (x) - \int \frac{x^{2} - a^{2} + a^{2}}{\sqrt{x^{2} - a^{2}}} d x$
$= \sqrt{x^{2} - a^{2}} (x) - \int \frac{x^{2} - a^{2}}{\sqrt{x^{2} - a^{2}}} d x + a^{2} \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x$
$= \sqrt{x^{2} - a^{2}} (x) - \int \sqrt{x^{2} - a^{2}} d x + a^{2} \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x$
$2 \int \sqrt{x^{2} - a^{2}} d x = \sqrt{x^{2} - a^{2}} (x) + a^{2} \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x$
$\int \sqrt{x^{2} - a^{2}} d x = \frac{1}{2} \sqrt{x^{2} - a^{2}} (x) + \frac{1}{2} a^{2} log | x + \sqrt{x^{2} - a^{2}} | + c$ ....where $c$ is constant term.

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Q. Prove that :-

\sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} log ∣ ∣ x + \sqrt{x^{2} + a^{2}} ∣ ∣ + c

\int \frac{x}{\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} - a^{2}}} d x

\int \frac{1}{\sqrt{x^{2} + a^{2}}} d x = l o g | x + \sqrt{x^{2} + a^{2}} | + c

\int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} {sin}^{- 1} (\frac{x}{a}) + c

\int \frac{d x}{\sqrt{x^{2} - a^{2}}} = log (\frac{x + \sqrt{x^{2} - a^{2}}}{a}) + c

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