1
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
$\tan^{-1}x + \tan^{-1}(x^3) + c$
2
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
$\sqrt{6}\,\tan^{-1}\left(\dfrac{\sqrt{3}x}{\sqrt{2}}\right) + c$
3
MHT CET 2026 16th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\displaystyle\int \dfrac{(x + 1)(x + \log x)^2}{x}\,dx =$
A
$\left(\dfrac{x + \log x}{x}\right)^2 + c$, where c is the constant of integration
B
$\dfrac{(x + \log x)^2}{x} + c$, where c is the constant of integration
C
$\dfrac{(x + \log x)^3}{3} + c$, where c is the constant of integration
D
$\dfrac{(x + \log x)^3}{3x} + c$, where c is the constant of integration
4
MHT CET 2026 16th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\displaystyle\int \cot^4 x\,dx$ is equal to
A
$\dfrac{\cot^3 x}{3} - \cot x + x + c$
B
$\dfrac{\cot^3 x}{3} + \cot x + x + c$
C
$-\dfrac{\cot^3 x}{3} + \cot x + x + c$
D
$\dfrac{\cot^3 x}{3} - 2\cot x + x + c$

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