1
MHT CET 2026 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If $f(x) = \int \dfrac{x}{(x^2 + 4)(x^2 + 9)}\, dx$ and $f(0) = \dfrac{1}{5}\log\left(\dfrac{2}{3}\right)$, then $f(1) = $
A
$\dfrac{1}{5}\log(2)$
B
$\dfrac{1}{10}\log(2)$
C
$-\dfrac{1}{5}\log(2)$
D
$-\dfrac{1}{10}\log(2)$
2
MHT CET 2026 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\int \dfrac{x^4(x^{10} - 1)}{x^{20} + 3x^{10} + 1}\, dx = \cdots$
A
$\tan^{-1}\left(x^5 + \dfrac{1}{x^5}\right) + c$
B
$\dfrac{1}{5}\tan^{-1}\left(x^5 + \dfrac{1}{x^5}\right) + c$
C
$\tan^{-1}\left(x^{10} + \dfrac{1}{x^{10}}\right) + c$
D
$\dfrac{1}{10}\tan^{-1}\left(x^{10} + \dfrac{1}{x^{10}}\right) + c$
3
MHT CET 2026 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\int \left(e^{\log(\sin x)} + \cos x\right) x\, dx = $
A
$x(\sin x + \cos x) + (\sin x - \cos x) + c$
B
$x(\sin x - \cos x) + (\sin x - \cos x) + c$
C
$x(\sin x + \cos x) + (\sin x + \cos x) + c$
D
$x(\sin x - \cos x) + (\sin x + \cos x) + c$
4
MHT CET 2026 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
The value of $\int \sin 4x \cos 3x\, dx$ is
A
$-\dfrac{1}{14}\cos 7x - \dfrac{1}{2}\cos x + c$
B
$-\dfrac{1}{14}\cos 7x + \dfrac{1}{2}\cos x + c$
C
$\dfrac{1}{14}\cos 7x - \dfrac{1}{2}\cos x + c$
D
$\dfrac{1}{14}\cos 7x + \dfrac{1}{2}\cos x + c$

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