Explanation

(b)
$\int_{}^{}{\sqrt{x^{2} + a^{2}}\ dx} = \int_{}^{}{\sqrt{x^{2} + a^{2}} \bullet 1\ dx}$
$= \sqrt{x^{2} + a^{2}}\int_{}^{}{1dx - \int_{}^{}{\left\lbrack \frac{d}{dx}\left( \ \sqrt{x^{2} + a^{2}}\ \right)\int_{}^{}{1dx} \right\rbrack dx}}$
$= x\sqrt{x^{2} + a^{2}} - \int_{}^{}{\left\lbrack \frac{2x}{2\sqrt{x^{2} + a^{2}}}x \right\rbrack dx}$
$= x\sqrt{x^{2} + a^{2}} - \int_{}^{}\left\lbrack \frac{x^{2} + a^{2} - a^{2}}{\sqrt{x^{2} + a^{2}}} \right\rbrack dx$
$= x\sqrt{x^{2} + a^{2}} - \int_{}^{}{\left\lbrack \sqrt{x^{2} + a^{2}} - \frac{a^{2}}{\sqrt{x^{2} + a^{2}}} \right\rbrack dx}$
$= x\sqrt{x^{2} + a^{2}} - \int_{}^{}{\sqrt{x^{2} + a^{2}}dx + a^{2}\int_{}^{}\frac{dx}{\sqrt{x^{2} + a^{2}}}}$
$= x\sqrt{x^{2} + a^{2}} - I + a^{2}\log{\lbrack x + \sqrt{x^{2} + a^{2}}\rbrack} + c$
$\Rightarrow 2I = x\sqrt{x^{2} + a^{2}} + a^{2}\log{\lbrack x + \sqrt{x^{2} + a^{2}}\rbrack} + c$
$\Rightarrow I = \frac{x}{2}\sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2}\log{\left\lbrack \ x\sqrt{x^{2} + a^{2}}\ \right\rbrack + c}$