MHT CET 2023 13th May Evening Shift | Indefinite Integration Question 13 | Mathematics

If $$I=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x$$ and $$J=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x$$, then for any arbitrary constant $$C$$, than the value of $$J-I$$ equals

$$\frac{1}{2} \log \left|\left(\frac{e^{4 x}-e^{2 x}+1}{e^{4 x}+e^{2 x}+1}\right)\right|+C$$

$$\frac{1}{2} \log \left|\left(\frac{e^{2 x}+e^x+1}{e^{2 x}-e^x+1}\right)\right|+C$$

$$\frac{1}{2} \log \left|\left(\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right)\right|+C$$

$$\frac{1}{2} \log \left|\left(\frac{e^{4 x}+e^{2 x}+1}{e^{4 x}-e^{2 x}+1}\right)\right|+C$$

MHT CET 2023 13th May Morning Shift

MCQ (Single Correct Answer)

-0

If $$\mathrm{I}=\int \frac{2 x-7}{\sqrt{3 x-2}} \mathrm{~d} x$$, then $$\mathrm{I}$$ is given by

$$\frac{106}{27}(3 x-2)^{\frac{3}{2}}+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.

$$\frac{98}{27}(3 x-2)^{\frac{3}{2}}+\mathrm{c}$$, where c is a constant of integration.

$$\frac{4}{27}(3 x-2)^{\frac{3}{2}}-\frac{34}{9}(3 x-2)^{\frac{1}{2}}+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.

$$\frac{4}{27}(3 x-2)^{\frac{3}{2}}+\frac{34}{9}(3 x-2)^{\frac{1}{2}}+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration

MHT CET 2023 13th May Morning Shift

MCQ (Single Correct Answer)

-0

$$\int \frac{\log \left(x^2+a^2\right)}{x^2} d x=$$

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

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

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

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

MHT CET 2023 13th May Morning Shift

MCQ (Single Correct Answer)

-0

If $$\int x^5 e^{-4 x^3} \mathrm{~d} x=\frac{1}{48} \mathrm{e}^{-4 x^3} \mathrm{f}(x)+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration, then $$\mathrm{f}(x)$$ is given by

$$4 x^3+1$$

$$-4 x^3-1$$

$$-2 x^3-1$$

$$-2 x^3+1$$

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