MHT CET 2023 9th May Evening Shift | Indefinite Integration Question 105 | Mathematics

Let $$\alpha \in\left(0, \frac{\pi}{2}\right)$$ be fixed. If the integral $$\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} \mathrm{d} x=\mathrm{A}(x) \cos 2 \alpha+\mathrm{B}(x) \sin 2 \alpha+\mathrm{c},$$ (where $$\mathrm{c}$$ is a constant of integration), then functions $$\mathrm{A}(x)$$ and $$\mathrm{B}(x)$$ are respectively

$$x+\alpha$$ and $$\log |\sin (x+\alpha)|$$.

$$x-\alpha$$ and $$\log |\sin (x-\alpha)|$$.

$$x-\alpha$$ and $$\log |\cos (x-\alpha)|$$.

$$x+\alpha$$ and $$\log |\sin (x-\alpha)|$$.

MHT CET 2023 9th May Morning Shift

MCQ (Single Correct Answer)

-0

$$\int \frac{x+1}{x\left(1+x \mathrm{e}^x\right)^2} \mathrm{~d} x=$$

$$\log \left|\frac{x \mathrm{e}^x}{1+x \mathrm{e}^x}\right|+c$$, where $$\mathrm{c}$$ is a constant of integration.

$$\log \left|\frac{x \mathrm{e}^x}{1+x \mathrm{e}^x}\right|-\frac{1}{1+x \mathrm{e}^x}+\mathrm{c}$$, where c is a constant of integration.

$$\log \left|1+x \mathrm{e}^x\right|+\frac{1}{1+x \mathrm{e}^x}+\mathrm{c}$$, where $$\mathrm{c}$$ is constant of integration.

$$\log \left|\frac{x \mathrm{e}^x}{1+x \mathrm{e}^x}\right|+\frac{1}{1+x \mathrm{e}^x}+\mathrm{c}$$, where $$\mathrm{c}$$ is constant of integration.

MHT CET 2023 9th May Morning Shift

MCQ (Single Correct Answer)

-0

$$\int \frac{\mathrm{e}^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \mathrm{d} x, x > 0=$$

$$\left(\tan ^{-1} x\right)^2 \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.

$$\left(\tan ^{-1} x\right) \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.

$$\left(\tan ^{-1} x\right) \mathrm{e}^{2 \tan ^{-1} x}+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.

$$\left(\tan ^{-1} x\right)^2 \mathrm{e}^{2 \tan ^{-1} x}+c$$, where $$\mathrm{c}$$ is a constant of integration.

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