1
MHT CET 2026 16th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
If $f(x) = \cos x$, then the value of $\int\dfrac{e^{f(x)}(x\sin^3 x + f(x))}{1 - (f(x))^2}dx = $
A
$e^{f(x)}(-\text{cosec}\ x - x) + c$
B
$e^{f(x)}(\text{cosec}\,x + x) + c$
C
$e^{f(x)}(\cos x - x) + c$
D
$e^{f(x)}(\cot^2 x + x) + c$
2
MHT CET 2026 16th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
$\int\dfrac{x^4 + 1}{x^6 + 1}dx = $
A
$\tan^{-1}x - \dfrac{1}{3}\tan^{-1}(x^3) + c$
B
$\tan^{-1}x + \dfrac{1}{3}\tan^{-1}(x^3) + c$
C
$\tan^{-1}x - \tan^{-1}(x^3) + c$
D
$\tan^{-1}x + \tan^{-1}(x^3) + c$
3
MHT CET 2026 16th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
If $a > 0, b > 0$ and $\int\dfrac{1}{ax^2 + b}dx = \dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{2}x}{\sqrt{3}}\right) + c$, then $\int\dfrac{1}{bx^2 + a}dx = \ldots$
A
$-\dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{2}x}{\sqrt{3}}\right) + c$
B
$\dfrac{1}{\sqrt{6}}\tan^{-1}\left(\dfrac{\sqrt{3}x}{\sqrt{2}}\right) + c$
C
$-\sqrt{6}\,\tan^{-1}\left(\dfrac{\sqrt{2}x}{\sqrt{3}}\right) + c$
D
$\sqrt{6}\,\tan^{-1}\left(\dfrac{\sqrt{3}x}{\sqrt{2}}\right) + c$
4
MHT CET 2026 16th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
The value of integral $\int_0^1\cot^{-1}(1 + x^2 - x)\,dx$ is...
A
$\dfrac{\pi}{2} - \log 2$
B
$\pi - \log 2$
C
$\dfrac{\pi}{4} - \log 2$
D
$2\pi - \log 2$

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