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Integrals

Multiple Choice Questions

721.

$\int \frac{dx}{x^{2} + 4 x + 13}$ is equal to

$\log (x^{2} + 4 x + 13) + c$
$\frac{1}{3} \tan^{- 1} (\frac{x + 2}{3}) + c$
$\log (2 x + 4) + c$
$\frac{2 x + 4}{{(x^{2} + 4 x + 13)}^{2}} + c$

722.

The value of $\int_{2}^{3} \frac{x + 1}{x^{2} (x - 1)} dx$ is

$\log (\frac{16}{9}) + \frac{1}{6}$
$\log (\frac{16}{9}) - \frac{1}{6}$
$2 \log (2) - \frac{1}{6}$
$\log (\frac{4}{3}) - \frac{1}{6}$

723.

$\int_{0}^{\frac{π}{4}} (\cos (x) - \sin (x)) dx + \int_{\frac{π}{4}}^{\frac{5 π}{4}} (\sin (x) - \cos (x)) dx + \int_{2 π}^{\frac{π}{4}} (\cos (x) - \sin (x)) dx$

is equal to

$\sqrt{2} - 2$
$2 \sqrt{2} - 2$
$3 \sqrt{2} - 2$
$4 \sqrt{2} - 2$

724.

$\int \frac{a^{\frac{x}{2}}}{\sqrt{a^{- x} - a^{x}}} dx$ is equal to

$\frac{1}{\log (a)} \sin^{- 1} (a^{x}) + c$
$\frac{1}{\log (a)} \tan^{- 1} (a^{x}) + c$
$2 \sqrt{a^{- x} - a^{x}} + c$
$\log (a^{x} - 1) + c$

725.

The value of $\int_{0}^{1} \frac{x^{4} + 1}{x^{2} + 1} d x$ is

$\frac{1}{6} (3 - 4 π)$
$\frac{1}{6} (3 π + 4)$
$\frac{1}{6} (3 + 4 π)$
$\frac{1}{6} (3 π - 4)$

726.

$\int \frac{e^{x}}{(2 + e^{x}) (e^{x} + 1)} dx$ is equal to

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

727.

$\int 32 x^{3} {(\log (x))}^{2} dx$ is equal to

8x⁴(log(x))² + c
$x^{4} \{8 {(\log (x))}^{2} - 4 (\log (x)) + 1\} + c$
$\{8 {(\log (x))}^{2} - 4 (\log (x))\} + c$
$x^{3} \{8 {(\log (x))}^{2} - 2 (\log (x))\} + c$

728.

$\int \frac{\cos (x) - 1}{\sin (x) + 1} e^{x} dx$ is equal to :

$\frac{e^{x} \cos (x)}{1 + \sin (x)} + c$
$c - \frac{e^{x} \sin (x)}{1 + \sin (x)}$
$c - \frac{e^{x}}{1 + \sin (x)}$
$c - \frac{e^{x} \cos (x)}{1 + \sin (x)}$

729.
If $\int f (x) dx = g (x) + c, then \int f^{- 1} (x) dx$ is equal to :

xf^-1(x) + c

f(g^-1(x)) + c

xf^-1(x) - g(f^-1(x)) + c

g^-1(x) + c

xf^-1(x) - g(f^-1(x)) + c

$Let I = \int f^{- 1} (x) dx . . . (i) and \int f (x) dx = g (x) + c . . . (ii) ∴ From (i) let f^{- 1} (x) = u \Rightarrow x = f (u) \Rightarrow dx = f' (u) du ∴ I = \int uf' (u) du Integration by parts, we get I = uf (u) - \int f (u) du I = uf (u) - g (u) + c [∵ usin Eq . (ii)] ∴ On putting u = f^{- 1} (x), f (u) = x We get, I = x f^{- 1} (x) - g (f^{- 1} (x)) + c$