The integral I = ∫ e e + 1 1 + x 2 1 + x 3 d x satisfies, I > 2, I > e, I 0, I 1

x ∈ e , e + 1 ⇒ x 2 x 3 ⇒ 1 + x 2 1 + x 3 1 ⇒ ∫ e e + 1 1 + x 2 1 + x 3 d x ∫ e e + 1 1 d x ⇒ I x e e + 1 ⇒ I 1

The integral I = ∫ e e + 1 1 + x 2 1 + x 3 d x satisfies

Question: Answered & Verified by Expert

The integral

I = \int_{e}^{e + 1} \frac{1 + x^{2}}{1 + x^{3}} d x

satisfies

MathematicsDefinite IntegrationJEE Main

Options:

Solution:

2749 Upvotes Verified Answer

The correct answer is:

I < 1

x \in (e, e + 1)

\Rightarrow x^{2} < x^{3}

\Rightarrow \frac{1 + x^{2}}{1 + x^{3}} < 1

\Rightarrow \int_{e}^{e + 1} \frac{1 + x^{2}}{1 + x^{3}} d x < \int_{e}^{e + 1} 1 d x

\Rightarrow I < {[x]}_{e}^{e + 1}

\Rightarrow I < 1

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