₀^{ 5} {e^{x} {e^{x}-1}}{e^{x}+3} d x=

Correct Answer is : 4−π4-π4−π

Correct Answer is : 4−π4-π4−π

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MHTCET 16 Apr 2026 Shift 1 Paper

Question : 130 of 150

Marks: +1, -0

\int_{0}^{\log 5} \frac{e^{x} \sqrt{e^{x}-1}}{e^{x}+3} d x=

Solution:

Concept:

Use substitution to simplify the integrand by setting

t = \sqrt{e^x - 1}

and then evaluate the resulting standard integral.

Explanation:

Let

t = \sqrt{e^x - 1}

Then

t^2 = e^x - 1

, so

e^x = t^2 + 1

Differentiating:

e^x\,dx = 2t\,dt

Change the limits:

When

x = 0

t = \sqrt{1 - 1} = 0

When

x = \log 5

t = \sqrt{5 - 1} = 2

Also,

e^x + 3 = t^2 + 4

Therefore, the integral becomes:

\int_{0}^{\log 5} \frac{e^{x} \sqrt{e^{x}-1}}{e^{x}+3}\,dx = \int_{0}^{2} \frac{t}{t^2 + 4} \cdot 2t\,dt = \int_{0}^{2} \frac{2t^2}{t^2 + 4}\,dt.

Simplify the integrand:

\frac{2t^2}{t^2 + 4} = 2 - \frac{8}{t^2 + 4}.

Now integrate:

\int_{0}^{2} \left(2 - \frac{8}{t^2 + 4}\right)dt = \left[2t - 4\tan^{-1}\left(\frac{t}{2}\right)\right]_{0}^{2}.

Evaluating at the limits:

= 4 - 4\tan^{-1}(1) = 4 - 4 \cdot \frac{\pi}{4} = 4 - \pi.

Answer:

4 - \pi

So, the correct option is D.

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