Question

\(\int_0^2 \frac{3x+1}{x^2+4} dx\)

• A. \(\log(2\sqrt{2}) + \pi/4\)
• B. \(\log(2\sqrt{2}) + \pi/6\)
• C. \(\log(2\sqrt{2}) + \pi/8\)
• D. \(\log(2\sqrt{2}) + \pi/12\)

Hint:

Whenever you see \(\frac{ax+b}{x^2+c}\), split it into: \[ \frac{ax}{x^2+c}+\frac{b}{x^2+c} \] because each part integrates in a standard way.

Answer 1

The Correct Option is C

Concept:
Split the integral into two standard forms:
• \(\int \frac{x}{x^2+a^2}\,dx\)
• \(\int \frac{dx}{x^2+a^2}\) e} ip

Step 1: Split the integrand.
\[ \int_0^2 \frac{3x+1}{x^2+4}\,dx = \int_0^2 \frac{3x}{x^2+4}\,dx + \int_0^2 \frac{1}{x^2+4}\,dx \] ip

Step 2: Evaluate the first integral.
\[ \int_0^2 \frac{3x}{x^2+4}\,dx = \frac{3}{2}\int_0^2 \frac{2x}{x^2+4}\,dx \] \[ = \frac{3}{2}\left[\ln(x^2+4)\right]_0^2 \] \[ = \frac{3}{2}(\ln 8-\ln 4) = \frac{3}{2}\ln 2 \] \[ \frac{3}{2}\ln 2 = \ln(2\sqrt{2}) \] ip

Step 3: Evaluate the second integral.
\[ \int_0^2 \frac{1}{x^2+4}\,dx = \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)\Bigg|_0^2 \] \[ = \frac12\left(\tan^{-1}1-\tan^{-1}0\right) = \frac12\cdot\frac{\pi}{4} = \frac{\pi}{8} \] ip

Step 4: Add both parts.
\[ \int_0^2 \frac{3x+1}{x^2+4}\,dx = \log(2\sqrt{2})+\frac{\pi}{8} \] ip Hence, the correct answer is:
\[ \boxed{(C)\ \log(2\sqrt{2})+\frac{\pi}{8}} \]