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. \(e^{\frac{1}{\sqrt t}}+C\)

Hint:

We want to evaluate the integral: \[\int \frac{e^{\frac{1}{\sqrt{t}}}}{t\sqrt{t}} dt\]

Let's use the substitution method. Let \(u = \frac{1}{\sqrt{t}} = t^{-\frac{1}{2}}\). Then, we find the derivative of \(u\) with respect to \(t\): \[\frac{du}{dt} = -\frac{1}{2}t^{-\frac{3}{2}} = -\frac{1}{2t\sqrt{t}}\]

Now, we can solve for \(dt\): \[dt = -2t\sqrt{t} \, du\]

Substitute \(u\) and \(dt\) into the integral: \[\int \frac{e^{\frac{1}{\sqrt{t}}}}{t\sqrt{t}} dt = \int \frac{e^u}{t\sqrt{t}} (-2t\sqrt{t}) \, du = \int -2e^u \, du\]

Now, we can integrate with respect to \(u\): \[\int -2e^u \, du = -2e^u + C\]

Finally, substitute back \(u = \frac{1}{\sqrt{t}}\): \[-2e^{\frac{1}{\sqrt{t}}} + C\]

Therefore, the integral is: \[\int \frac{e^{\frac{1}{\sqrt{t}}}}{t\sqrt{t}} dt = -2e^{\frac{1}{\sqrt{t}}} + C\]

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) : \(-2e^{\frac{1}{\sqrt t}}+C\)