Morgan Weston
01/24/2024 · Junior High School
Compute \( \int \frac{2^{\sqrt{3 x+1}}}{\sqrt{3 x+1}} d x \)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To compute the integral
\[
I = \int \frac{2^{\sqrt{3x+1}}}{\sqrt{3x+1}} \, dx,
\]
we can use a substitution to simplify the expression. Let
\[
u = \sqrt{3x + 1}.
\]
Then, we differentiate \( u \) with respect to \( x \):
\[
u^2 = 3x + 1 \implies 2u \frac{du}{dx} = 3 \implies \frac{du}{dx} = \frac{3}{2u} \implies dx = \frac{2u}{3} \, du.
\]
Next, we express \( x \) in terms of \( u \):
\[
3x = u^2 - 1 \implies x = \frac{u^2 - 1}{3}.
\]
Now, we substitute \( dx \) and \( u \) into the integral:
\[
I = \int \frac{2^u}{u} \cdot \frac{2u}{3} \, du = \frac{2}{3} \int 2^u \, du.
\]
The integral of \( 2^u \) is given by:
\[
\int 2^u \, du = \frac{2^u}{\ln 2} + C.
\]
Thus, we have:
\[
I = \frac{2}{3} \cdot \left( \frac{2^u}{\ln 2} \right) + C = \frac{2^{\sqrt{3x+1}}}{3 \ln 2} + C.
\]
Finally, we can write the result of the integral as:
\[
\int \frac{2^{\sqrt{3x+1}}}{\sqrt{3x+1}} \, dx = \frac{2^{\sqrt{3x+1}}}{3 \ln 2} + C.
\]
Quick Answer
\[
\int \frac{2^{\sqrt{3x+1}}}{\sqrt{3x+1}} \, dx = \frac{2^{\sqrt{3x+1}}}{3 \ln 2} + C.
\]
Answered by UpStudy AI and reviewed by a Professional Tutor
UpStudy ThothAI
Self-Developed and Ever-Improving
Thoth AI product is constantly being upgraded and optimized.
Covers All Major Subjects
Capable of handling homework in math, chemistry, biology, physics, and more.
Instant and Accurate
Provides immediate and precise solutions and guidance.
Try Now
Ask Tutors
Ask AI
10x
Fastest way to Get Answers & Solutions
By text
Enter your question here…
By image
Re-Upload
Submit