∫dxxx7 + 1 is equal to from Mathematics Integ

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Integrals

Multiple Choice Questions

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741.
$\int \frac{dx}{x (x^{7} + 1)}$ is equal to

$\log (\frac{x^{7}}{x^{7} + 1}) + c$

$\frac{1}{7} \log (\frac{x^{7}}{x^{7} + 1}) + c$

$\log (\frac{x^{7} + 1}{x^{7}}) + c$

$\frac{1}{7} \log (\frac{x^{7} + 1}{x^{7}}) + c$

B.

$\frac{1}{7} \log (\frac{x^{7}}{x^{7} + 1}) + c$

$Given that, \int \frac{dx}{x (x^{7} + 1)} On putting x^{7} = t \Rightarrow 7 x^{6} dx = dt \Rightarrow dx = \frac{1}{7 x^{6}} dt ∴ I = \int \frac{1}{7 x^{7}} \frac{dt}{(t + 1)} = \frac{1}{7} \int \frac{1}{t} \frac{dt}{(t + 1)} [∵ x^{7} = t] = \frac{1}{7} \int (\frac{1}{t} - \frac{1}{t + 1}) = \frac{1}{7} \log (\frac{t}{t + 1}) + c = \frac{1}{7} \log (\frac{x^{7}}{x^{7} + 1}) + c$

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742.

$\int_{- 1}^{1} |1 - x| dx$ is equal to

- 2
0
2
4

743.

$\int \sqrt{x} e^{\sqrt{x}} dx$ is equal to

$2 \sqrt{x} - e^{\sqrt{x}} - 4 \sqrt{{xe}^{\sqrt{x}}} + c$
$(2 \sqrt{x} - 4 \sqrt{x} + 4) e^{\sqrt{x}} + c$
$(2 \sqrt{x} + 4 \sqrt{x} + 4) e^{\sqrt{x}} + c$
$(1 - 4 \sqrt{x}) e^{\sqrt{x}} + c$

744.

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

$\sin^{- 1} (x + 1) + c$
$\sinh^{- 1} (x + 1) + c$
$\tanh^{- 1} (x + 1) + c$
$\tan^{- 1} (x + 1) + c$

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745.

$\int_{0}^{2 π} (\sin (x) + |\sin (x)|) dx$ is equal to

4
0
1
8

746.

$\int_{0}^{π} \frac{xdx}{a^{2} \cos^{2} (x) + b^{2} \sin^{2} (x)} dx$ is equal to

$\frac{π}{2 ab}$
$\frac{π}{ab}$
$\frac{π^{2}}{2 ab}$
$\frac{π^{2}}{ab}$

747.

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

$e^{x} \sec^{2} (\frac{x}{2}) + c$
$e^{x} \tan (\frac{x}{2})$ + c
$e^{x} \sec (\frac{x}{2}) + c$
$e^{x} \tan (x) + c$

748.

$\int \sqrt{1 + \sin (\frac{x}{4})} dx$ is equal to

$8 (\sin (\frac{x}{8}) + \cos (\frac{x}{8})) + C$
$8 (\sin (\frac{x}{8}) - \cos (\frac{x}{8})) + C$
$8 (\cos (\frac{x}{8}) - \sin (\frac{x}{8})) + C$
$\frac{1}{8} (\sin (\frac{x}{8}) - \cos (\frac{x}{8})) + C$

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749.

$\int_{0}^{\infty} \frac{xdx}{(1 + x) (1 + x^{2})}$ is equal to

$\frac{π}{2}$
0
1
$\frac{π}{4}$

750.

If I_n = $\int {(\log (x))}^{n} dx$ , then I_n + nI_{n - 1} is equal to

${(xlog (x))}^{n}$
$x {(\log (x))}^{n}$
$n {(\log (x))}^{n}$
${(\log (x))}^{n - 1}$

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