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Integrals

Multiple Choice Questions

451.

$\int (\sqrt[3]{x}) (\sqrt[5]{1 + \sqrt[3]{x^{4}}}) dx$ is equal to

${(1 + x^{\frac{3}{4}})}^{\frac{6}{5}} + c$
${(1 + x^{\frac{4}{3}})}^{\frac{6}{5}} + c$
$\frac{5}{8} {(1 + x^{\frac{4}{3}})}^{\frac{6}{5}} + c$
$\frac{1}{6} {(1 + x^{\frac{4}{3}})}^{6} + c$

452.

If u = - $f'' (θ) \sin (θ) + f' (θ) \cos (θ)$ and v = $f'' (θ) \cos (θ) + f' (θ) \sin (θ)$ , then $\int {[{(\frac{du}{dθ})}^{2} + {(\frac{dv}{dθ})}^{2}]}^{\frac{1}{2}} dθ$ is equal to

$f (θ) - f'' (θ) + c$
$f (θ) + f'' (θ) + c$
$f' (θ) + f'' (θ) + c$
$f' (θ) - f'' (θ) + c$

453.

$\int \frac{e^{6 \log_{e} (x)} - e^{5 \log_{e} (x)}}{e^{4 \log_{e} (x)} - e^{3 \log_{e} (x)}} dx$ is equal to

$\frac{x^{3}}{3} + c$
$\frac{x^{2}}{2} + c$
$\frac{x^{2}}{3} + c$
$\frac{- x^{3}}{3} + c$

454.

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

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

455.

$\int \frac{x^{4} - 1}{x^{2} {(x^{4} + x^{2} + 1)}^{\frac{1}{2}}} dx$ is equal to

$\sqrt{\frac{x^{4} + x^{2} + 1}{x}} + c$
$\frac{x^{2}}{\sqrt{x^{4} + x^{2} + 1}} + c$
$x {(x^{4} + x^{2} + 1)}^{\frac{3}{2}} + c$
$\frac{\sqrt{x^{4} + x^{2} + 1}}{x} + c$

456.

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

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

457.

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

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

458.

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

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

459.

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

460.
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

$\frac{1}{4}$

Let x + 2 = $\frac{d}{dx} (2 x^{2} + 6 x + 5) + B$

A $\Rightarrow x + 2 = A (4 x + 6) + B \Rightarrow x + 2 = 4 Ax + 6 A + B \Rightarrow 4 A = 1 \Rightarrow A = \frac{1}{4} 6 A + B = 2 \Rightarrow B = \frac{1}{2} ∴ \int \frac{x + 2}{2 x^{2} + 6 x + 5} dx = \int \frac{(\frac{1}{4} (4 x + 6) + \frac{1}{2})}{2 x^{2} + 6 x + 5} dx = \frac{1}{4} \int \frac{4 x + 6}{2 x^{2} + 6 x + 5} dx + \frac{1}{2} \int \frac{1}{2 x^{2} + 6 x + 5} dx comparing it with \int \frac{x + 2}{2 x^{2} + 6 x + 5} dx = P \int \frac{4 x + 6}{2 x^{2} + 6 x + 5} dx + \frac{1}{2} \int \frac{dx}{2 x^{2} + 6 x + 5} \Rightarrow P = \frac{1}{4}$

Previous Year Papers

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Multiple Choice Questions

460. If ∫x + 22x2 + 6x + 5dx = P ∫4x + 62x2 + 6x + 5dx + 12∫dx2x2 + 6x + 5, then the values of P is 13 12 14 2

460.
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