1
MHT CET 2026 15th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\int \dfrac{\sqrt{x-2}}{x}dx = $
A
$2\sqrt{x-2} - 2\sqrt{2}\tan^{-1}\left(\dfrac{\sqrt{x-2}}{\sqrt{2}}\right) + c$
B
$2\sqrt{x-2} + 2\sqrt{2}\tan^{-1}\left(\dfrac{\sqrt{x-2}}{\sqrt{2}}\right) + c$
C
$2\sqrt{x-2} - 2\sqrt{2}\tan^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$
D
$2\sqrt{x-2} + 2\sqrt{2}\tan^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$
2
MHT CET 2026 15th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If $f(x)$ and $g(x)$ are integrable functions then $\left[\int f(x)dx\right]\left[\int g(x)dx\right] = $
A
$\int \left[f(x)g'(x) + f'(x)g(x)\right]dx$
B
$\int \left[f(x)g'(x) - f'(x)g(x)\right]dx$
C
$\int \left[f(x)\int g(x)dx + g(x)\int f(x)dx\right]dx$
D
$\int \left[f(x)\int g(x)dx - g(x)\int f(x)dx\right]dx$
3
MHT CET 2026 15th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If an antiderivative of $f(x)$ is $e^x$ and an antiderivative of $g(x)$ is $\cos x$, then $\int f(x)\cos x\, dx + \int g(x)e^x\, dx = $
A
$e^x \sin x + c$
B
$e^x(f(x) + g(x)) + c$
C
$e^x \cos x + c$
D
$e^x + c$
4
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

$$ \int \frac{d x}{\cos x(1+\cos x)}= $$

A

$\quad \log (\sec x+\tan x)+2 \tan \left(\frac{x}{2}\right)+\mathrm{c}$, where c is the constant of integration

B

$\quad \log (\sec x+\tan x)-2 \tan \left(\frac{x}{2}\right)+\mathrm{c}$, where c is the constant of integration

C

$\log (\sec x+\tan x)+\tan \left(\frac{x}{2}\right)+\mathrm{c}$, where c is the constant of integration

D

$\log (\sec x+\tan x)-\tan \left(\frac{x}{2}\right)+\mathrm{c}$, where c is the constant of integration

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