Find the integral.
-$\int \frac{3 \mathrm{e}^{\sqrt{\mathrm{z}}}}{8 \sqrt{\mathrm{z}}} \mathrm{dz}$
A) $-12 e^{\sqrt{z}}+C$
B) $-24 e^{\sqrt{z}}+C$
C) $\frac{3}{4} \mathrm{e}^{\sqrt{\mathrm{z}}}+\mathrm{C}$
D) $\frac{3}{8} e^{\sqrt{z}}+C$

Question

Quizplus · Accepted Answer

The integral can be solved by substitution. Let $ u = \sqrt{z} $, then $ du = \frac{1}{2\sqrt{z}} dz $ or $ dz = 2\sqrt{z} du $. Substitute and simplify to get $\int \frac{3 e^u}{8} \cdot 2 du = \frac{3}{4} \int e^u du = \frac{3}{4} e^u + C = \frac{3}{4} e^{\sqrt{z}} + C$.

Find the Integral ∫3ez8zdz\int \frac{3 \mathrm{e}^{\sqrt{\mathrm{z}}}}{8 \sqrt{\mathrm{z}}} \mathrm{dz}∫8z​3ez​​dz A) −12ez+C-12 e^{\sqrt{z}}+C−12ez​+C B) −24ez+C-24 e^{\sqrt{z}}+C−24ez​+C

Find the Integral $\int \frac{3 \mathrm{e}^{\sqrt{\mathrm{z}}}}{8 \sqrt{\mathrm{z}}} \mathrm{dz}$ A) $-12 e^{\sqrt{z}}+C$ B) $-24 e^{\sqrt{z}}+C$