1
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
2
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
$\dfrac{\cot^3 x}{3} - 2\cot x + x + c$
3
MHT CET 2026 16th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\displaystyle\int \dfrac{\cos^3 x}{\sin^2 x + \sin x}\,dx =$
A
$\log(\cos x) + \cos x + c$, where c is the constant of integration
B
$\log(\sin x) - \sin x + c$, where c is the constant of integration
C
$\log(\sin x) + \sin x + c$, where c is the constant of integration
D
$\log(\cos x) - \cos x + c$, where c is the constant of integration
4
MHT CET 2026 16th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\displaystyle\int_{\pi/6}^{\pi/3} \dfrac{\sin x - \cos x}{1 + \sin x\cos x}\,dx =$
A
$0$
B
$\dfrac{\pi}{12}$
C
$\dfrac{\pi}{24}$
D
$\dfrac{\pi^2}{2}$

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