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Integrals

Multiple Choice Questions

391.

$\int (\frac{4 e^{x} - 25}{2 e^{x} - 5}) dx = Ax + Blog (2 e^{x} - 5) + c$ , then

A = 5 and B = 3
A = 5 and B = - 3
A = - 5 and B = 3
A = - 5 and B = - 3

392.

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \log (\frac{2 - \sin (x)}{2 + \sin (x)}) dx$ is equal to

393.

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

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

394.

If $\int \frac{f (x)}{\log (\sin (x))} dx = \log [\log (\sin (x))] + c$ , then f(x) is equal to

cot(x)
tan(x)
sec(x)
csc(x)

395.

$\int_{0}^{\frac{π}{2}} (\frac{\sqrt[n]{\sec (x)}}{\sqrt[n]{\sec (x)} + \sqrt[n]{\csc (x)}}) dx$ is equal to

$\frac{π}{2}$
$\frac{π}{3}$
$\frac{π}{4}$
$\frac{π}{6}$

396.

$\int_{0}^{1} {xtan}^{- 1} (x) dx$ =

$\frac{π}{4} + \frac{1}{2}$
$\frac{π}{4} - \frac{1}{2}$
$\frac{1}{2} - \frac{π}{4}$
$- \frac{π}{4} - \frac{1}{2}$

397.

If $\int \frac{1}{\sqrt{9 - 16 x^{2}}} dx = {αsin}^{- 1} (βx) + c$ , then $α + \frac{1}{β}$ =

1
$\frac{7}{12}$
$\frac{19}{12}$
$\frac{9}{12}$

398.

If $\int_{0}^{\frac{π}{2}} \log (\cos (x)) dx = \frac{π}{2} \log (\frac{1}{2})$ , then $\int_{0}^{\frac{π}{2}} \log (\sec (x)) dx$ =

$\frac{π}{2} \log (\frac{1}{2})$
$1 - \frac{π}{2} \log (\frac{1}{2})$
$1 + \frac{π}{2} \log (\frac{1}{2})$
$\frac{π}{2} \log (2)$

399.

$\int \frac{1}{(x^{2} + 4) (x^{2} + 9)} dx = {Atan}^{- 1} (\frac{x}{2}) + {Btan}^{- 1} (\frac{x}{3}) + C$ , then A - B =

$\frac{1}{6}$
$\frac{1}{30}$
$- \frac{1}{30}$
$- \frac{1}{6}$

400.
If $\sqrt{\frac{x - 5}{x - 7}} dx = A \sqrt{x^{2} - 12 x + 35}$ + $\log |x - 6 + \sqrt{x^{2} - 12 x + 35}| + C$ , then A =

- 1

$\frac{1}{2}$

$- \frac{1}{2}$

1

$Let I = \int \sqrt{\frac{x - 5}{x - 7}} dx = \int \sqrt{\frac{(x - 5) (x - 5)}{(x - 7) (x - 5)}} dx [∵ rationalising] = \int \frac{x - 5}{\sqrt{x^{2} - 5 x - 7 x + 35}} dx = \int \frac{x - 5}{\sqrt{x^{2} - 12 x + 35}} = \frac{1}{2} \int \frac{2 x - 10}{\sqrt{x^{2} - 12 x + 35}} dx = \frac{1}{2} \int \frac{2 x - 12 + 2}{\sqrt{x^{2} - 12 x + 35}} dx = \frac{1}{2} \int \frac{2 x - 12}{\sqrt{x^{2} - 12 x + 35}} dx + \int \frac{1}{\sqrt{x^{2} - 12 x + 35}} dx = \sqrt{x^{2} - 12 x + 35} + \int \frac{1}{\sqrt{x^{2} - 12 x + 36 - 1}} + C [∵ Let x^{2} - 12 x + 35 = t \Rightarrow (2 x - 12) dx = dt] = \sqrt{x^{2} - 12 x + 35} + \int \frac{1}{\sqrt{{(x - 6)}^{2} - 1^{2}}} dx + C = \sqrt{x^{2} - 12 x + 35} + \log |x - 6 + \sqrt{x^{2} - 12 x + 35}| + C ∴ I = A \sqrt{x^{2} - 12 x + 35} + \log |x - 6 + \sqrt{x^{2} - 12 x + 35}| + C = 1 . \sqrt{x^{2} - 12 x + 35} + \log |x - 6 + \sqrt{x^{2} - 12 x + 35}| + C On comparing both sides, we get A = 1 .$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

400. If x - 5x - 7dx = Ax2 - 12x + 35 + logx - 6 + x2 - 12x + 35 + C, then A = - 1 12 - 12 1

400.
If $\sqrt{\frac{x - 5}{x - 7}} dx = A \sqrt{x^{2} - 12 x + 35}$ + $\log |x - 6 + \sqrt{x^{2} - 12 x + 35}| + C$ , then A =

- 1

$\frac{1}{2}$

$- \frac{1}{2}$

1