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Integrals

Multiple Choice Questions

991.
$\int_{\frac{π}{4}}^{\frac{π}{4}} (\frac{x + \frac{π}{4}}{2 - \cos (2 x)}) dx$ is equal to

$\frac{8 π \sqrt{3}}{5}$

$\frac{2 π \sqrt{3}}{9}$

$\frac{4 π^{2} \sqrt{3}}{9}$

$\frac{π^{2}}{6 \sqrt{3}}$

$\frac{π^{2}}{6 \sqrt{3}}$

$Let I = \int_{- \frac{π}{4}}^{\frac{π}{4}} (\frac{x + \frac{π}{4}}{2 - \cos (2 x)}) dx = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x}{2 - \cos (2 x)} + \frac{π}{4} \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{dx}{2 - \cos (2 x)} = 0 + 2 \times \frac{π}{4} \int_{0}^{\frac{π}{4}} \frac{dx}{2 - \frac{1 - \tan^{2} (x)}{1 + \tan^{2} (x)}} 0 = \frac{π}{2} \int_{0}^{\frac{π}{4}} \frac{1 + \tan^{2} (x)}{2 + 2 \tan^{2} (x) - 1 + \tan^{2} (x)} dx = \frac{π}{2} \int_{0}^{\frac{π}{4}} \frac{\sec^{2} (x)}{1 + 3 \tan^{2} (x)} dx Put \tan (x) = t \Rightarrow \sec^{2} (x) dx = dt ∴ \frac{π}{2} \int_{0}^{1} \frac{dt}{1 + 3 t^{2}} = \frac{π}{2} \times \frac{1}{\sqrt{3}} {[\tan^{- 1} (\sqrt{3} t)]}_{0}^{1} = \frac{π}{2 \sqrt{3}} [\tan^{- 1} (\sqrt{3}) - \tan^{- 1} (0)] = \frac{π}{2 \sqrt{3}} (\frac{π}{3}) = \frac{π^{2}}{6 \sqrt{3}}$

992.

$\int \frac{5 x + 3}{x^{2} (x^{2} - 2)} dx = ?$

$\frac{3}{2 x} + \frac{\sqrt{13}}{2 \sqrt{2}} \log (|\frac{\sqrt{2} - x}{\sqrt{2} + x}|) + C$
$\frac{3}{2 x} + \frac{13}{4 \sqrt{2}} \log (|\frac{x + \sqrt{2}}{x - \sqrt{2}}|) + C$
$\frac{3}{2 x} + \frac{13}{4 \sqrt{2}} \log (|\frac{x - \sqrt{2}}{x + \sqrt{2}}|) + C$
$\frac{3}{5} x + \frac{5}{3 \sqrt{2}} \log (|\frac{x + \sqrt{2}}{x - \sqrt{2}}|) + C$