QuestionJuly 15, 2025

Evaluate the indefinite integral given below. int (-4e^sqrt (4x))/(sqrt (4x))dx

Evaluate the indefinite integral given below. int (-4e^sqrt (4x))/(sqrt (4x))dx
Evaluate the indefinite integral given below. int (-4e^sqrt (4x))/(sqrt (4x))dx

Solution
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Answer

-8e^{\sqrt{4x}} + C Explanation 1. Substitution Let u = \sqrt{4x}, then du = \frac{1}{2\sqrt{4x}}dx. Thus, dx = 2\sqrt{4x} \, du = 2u \, du. 2. Rewrite the Integral Substitute u and dx into the integral: \int \frac{-4e^u}{u} \cdot 2u \, du = -8 \int e^u \, du. 3. Integrate The integral of e^u is e^u, so -8 \int e^u \, du = -8e^u + C. 4. Back-substitute Replace u with \sqrt{4x}: -8e^{\sqrt{4x}} + C.

Explanation

1. Substitution<br /> Let $u = \sqrt{4x}$, then $du = \frac{1}{2\sqrt{4x}}dx$. Thus, $dx = 2\sqrt{4x} \, du = 2u \, du$.<br /><br />2. Rewrite the Integral<br /> Substitute $u$ and $dx$ into the integral: $\int \frac{-4e^u}{u} \cdot 2u \, du = -8 \int e^u \, du$.<br /><br />3. Integrate<br /> The integral of $e^u$ is $e^u$, so $-8 \int e^u \, du = -8e^u + C$.<br /><br />4. Back-substitute<br /> Replace $u$ with $\sqrt{4x}$: $-8e^{\sqrt{4x}} + C$.
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