Question

$\int\frac{e^{\frac{1}{\sqrt t}}}{t\sqrt t}dt=$

• A. $\frac{1}{2}e^{\frac{1}{\sqrt t}}+C$
• B. $\frac{-1}{2}e^{\frac{1}{\sqrt t}}+C$
• C. $2e^{\frac{1}{\sqrt t}}+C$
• D. $-2e^{\frac{1}{\sqrt t}}+C$
• E. $e^{\frac{1}{\sqrt t}}+C$

Answer 1

The Correct Option is D

We are tasked with evaluating the integral: $$\int \frac{e^{\sqrt{t}}}{t \sqrt{t}} \, dt.$$
Step 1: Use substitution Let us perform the substitution: $$u = \sqrt{t} \quad \Rightarrow \quad t = u^2.$$ Now, differentiate both sides to find $dt$: $$dt = 2u \, du.$$ Substitute these into the integral: $$\int \frac{e^{\sqrt{t}}}{t \sqrt{t}} \, dt = \int \frac{e^u}{u^2 \cdot u} \cdot 2u \, du = \int \frac{2e^u}{u^2} \cdot u \, du.$$ Simplifying: $$= \int 2e^u \, du.$$
Step 2: Integrate The integral of $e^u$ with respect to $u$ is simply $e^u$. So we have: $$\int 2e^u \, du = 2e^u + C.$$
Step 3: Substitute back $u = \sqrt{t}$ Now, substitute $u = \sqrt{t}$ back into the expression: $$2e^u + C = 2e^{\sqrt{t}} + C.$$

The correct option is (D) : $-2e^{\frac{1}{\sqrt t}}+C$