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Integrals

Multiple Choice Questions

171.

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

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

172.

$\int \frac{1}{x} \log_{ex} (e) dx$ is equal to

$\log_{e} (1 - \log_{e} (x)) + c$
$\log_{e} (\log_{e} (ex) - 1)$ + c
$\log_{e} (\log_{e} (x) - 1)$ + c
$\log_{e} (1 + \log_{e} (x)) + c$

173.

The value of $\int_{1}^{e} 10^{\log_{e} (x)} d x$ is equal to

10 $\log_{e} (10 e)$
$\frac{10 e - 1}{\log_{e} (10 e)}$
$\frac{10 e}{\log_{e} (10 e)}$
$(10 e) \log_{e} (10 e)$

174.

The value of $\int_{- 2}^{4} |x + 1| dx$ is equal to

175.

If $\int \frac{x + 2}{2 x^{2} + 6 x + 5} dx$ = P $\int \frac{4 x + 6}{2 x^{2} + 6 x + 5} dx$ + $\frac{1}{2} \int \frac{dx}{2 x^{2} + 6 x + 5}$ , then the values of P is

$\frac{1}{3}$
$\frac{1}{2}$
$\frac{1}{4}$
2

176.

$\int (x + 1) {(x + 2)}^{7} (x + 3) dx$ is equal to

$\frac{{(x + 2)}^{10}}{10} - \frac{{(x + 2)}^{8}}{8} + c$
$\frac{{(x + 1)}^{10}}{2} - \frac{{(x + 2)}^{8}}{8} - \frac{{(x + 3)}^{2}}{2} + c$
$\frac{{(x + 2)}^{10}}{10} + c$
$\frac{{(x + 1)}^{2}}{2} + \frac{{(x + 2)}^{8}}{8} + \frac{{(x + 3)}^{2}}{2} + c$

177.

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

$\frac{{(x + 1)}^{\frac{7}{2}}}{7} - 2 \frac{{(x + 1)}^{\frac{5}{2}}}{5} + 2 \frac{{(x + 1)}^{\frac{3}{2}}}{3} + c$
$2 [\frac{{(x + 1)}^{\frac{7}{2}}}{7} - 2 \frac{{(x + 1)}^{\frac{5}{2}}}{5} + 2 \frac{{(x + 1)}^{\frac{3}{2}}}{3}] + c$
$\frac{{(x + 1)}^{\frac{7}{2}}}{7} - 2 \frac{{(x + 1)}^{\frac{5}{2}}}{5} + c$
${(x + 1)}^{\frac{7}{2}} + {(x + 1)}^{\frac{5}{2}} + {(x + 1)}^{\frac{3}{2}} + c$

178.

$\int \frac{1 + x}{x + e^{- x}} dx$ is equal to

$\log |(x - e^{- x})|$
$\log |(x + e^{- x})|$
$\log |(1 + {xe}^{x})| + c$
${(1 + {xe}^{x})}^{2}$ + c

179.
$\int \frac{\log (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}} dx$ is equal to

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

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

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

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

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

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