JEE Main Integration Practice Test 2025 - Free Integration Practice Questions and Study Guide

Question: 1 / 400

Evaluate ∫ (4/x^3) dx.

-2/x^2 + C

To evaluate the integral of \(\frac{4}{x^3}\) with respect to \(x\), we can rewrite the integrand in a more manageable form. The expression \(\frac{4}{x^3}\) can be expressed as \(4x^{-3}\).

Now, we can apply the power rule of integration, which states that \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) for \(n \neq -1\).

Here, \(n = -3\), so we apply the power rule:

\[

\int 4x^{-3} \, dx = 4 \cdot \frac{x^{-3 + 1}}{-3 + 1} + C = 4 \cdot \frac{x^{-2}}{-2} + C

\]

This simplifies to:

\[

4 \cdot \left(-\frac{1}{2} x^{-2}\right) + C = -2x^{-2} + C

\]

Reverting \(x^{-2}\) back to its original form \( \frac{1}{x^2} \), we find:

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2/x^2 + C

4/x^2 + C

-x^2 + C

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