₀^{ 10}≤ft[{e^{x} {e^{x}-1}}{e^{x}+8}] d x=

Correct Answer is : 32(4−π)3/2(4-π)23(4−π)

Correct Answer is : 32(4−π)3/2(4-π)23(4−π)

Test Index

MHTCET 15 Apr 2026 Shift 1 Paper

Question : 131 of 150

Marks: +1, -0

\int_{0}^{\log 10}\left[\frac{e^{x} \sqrt{e^{x}-1}}{e^{x}+8}\right] d x=

Solution:

Concept:

Substitution reduces the given integral to a standard inverse tangent form.

Explanation:

Let

I = \int_{0}^{\log 10} \frac{e^{x}\sqrt{e^{x}-1}}{e^{x}+8} dx

Put

t = \sqrt{e^{x}-1}

, so

t^{2} = e^{x}-1

and

e^{x} = t^{2}+1

Differentiating gives

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

Change the limits: when

x=0

t=0

; when

x=\log 10

t=3

Substituting,

I = \int_{0}^{3} \frac{2t^{2}}{t^{2}+9} dt

Rewrite

2t^{2}

2(t^{2}+9)-18

Thus

I = 2\int_{0}^{3} dt - 18\int_{0}^{3} \frac{dt}{t^{2}+9}

Since

\int \frac{dt}{t^{2}+9} = \frac{1}{3}\tan^{-1}\frac{t}{3}

, we get

I = 2[t]_{0}^{3} - 18\left[\frac{1}{3}\tan^{-1}\frac{t}{3}\right]_{0}^{3}

Therefore

I = 6 - 6\left(\tan^{-1}(1) - \tan^{-1}(0)\right) = 6 - 6\cdot\frac{\pi}{4}

This simplifies to

I = 6 - \frac{3\pi}{2} = \frac{3}{2}(4-\pi)

Answer:

Option B:

\frac{3}{2}(4-\pi)

Go to Question:

Prev Question

Next Question