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Integrals

Multiple Choice Questions

561.

$\int \sqrt{(\frac{1 - \sqrt{x}}{1 + \sqrt{x}})} dx$ is equal to

$\cos^{- 1} (\sqrt{x}) + \sqrt{1 - x} (\sqrt{x} - 2) + c$
$\cos^{- 1} (\sqrt{x}) - \sqrt{1 - x} (\sqrt{x} - 2) + c$
$\cos^{- 1} (\sqrt{x}) + \sqrt{1 - x} (\sqrt{x} + 2) + c$
None of the above

562.
$\int \frac{\log (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}} dx$ is equal to

$\frac{1}{2} {[\log (x + \sqrt{1 + x^{2}})]}^{2} + c$

$\log {(x + \sqrt{1 + x^{2}})}^{2} + c$

$\log (x + \sqrt{1 + x^{2}}) + c$

None of the above

$\frac{1}{2} {[\log (x + \sqrt{1 + x^{2}})]}^{2} + c$

$Let I = \int \frac{\log (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}} dx Put \log (x + \sqrt{1 + x^{2}}) = t \Rightarrow \frac{1}{\sqrt{1 + x^{2}}} dx = dt ∴ I = \int tdt = \frac{t^{2}}{2} + c = \frac{1}{2} {[\log (x + \sqrt{1 + x^{2}})]}^{2} + c$

563.

$\int \frac{\sin^{8} (x) - \cos^{8} (x)}{1 - 2 \sin^{2} (x) \cos^{2} (x)} dx$ is equal to

$\sin (2 x) + c$
$- \frac{1}{2} \sin (2 x) + c$
$\frac{1}{2} \sin (2 x) + c$
$- \sin (2 x) + c$

564.

$\int \frac{dx}{(\sin (x) + \sin (2 x))}$ is equal to

$\frac{1}{6} \log (1 - \cos (x)) + \frac{1}{2} \log (\log (1 + \cos (x))) - \frac{2}{3} \log (1 + 2 \cos (x)) + c$
$6 \log (1 - \cos (x)) + 2 \log (\log (1 + \cos (x))) - \frac{2}{3} \log (1 + 2 \cos (x)) + c$
$6 \log (1 - \cos (x)) + \frac{1}{2} \log (\log (1 + \cos (x))) + \frac{2}{3} \log (1 + 2 \cos (x)) + c$
None of the above