Question

The value of the integral \(\int\limits_2^4 \frac{x}{x^2+1} dx\) is:

• A. \(\frac{1}{2}log(\frac {5}{17})\)
• B. \(\frac{1}{2}log(\frac {17}{5})\)
• C. \(2log(\frac{5}{17})\)
• D. \(2log(\frac{17}{5})\)

Answer 1

The Correct Option is B

To solve the integral \(\int\limits_2^4 \frac{x}{x^2+1} dx\), we can use the substitution method. Let \(u = x^2 + 1\). Then \(du = 2x \, dx\) or \(x \, dx = \frac{1}{2} du\).

Next, we change the limits of integration accordingly. When \(x = 2\), \(u = 2^2 + 1 = 5\). When \(x = 4\), \(u = 4^2 + 1 = 17\). The integral becomes:

\(\int\limits_5^{17} \frac{1}{2} \frac{1}{u} \, du\).

This simplifies to \(\frac{1}{2} \int\limits_5^{17} \frac{1}{u} \, du\), which evaluates to:

\(\frac{1}{2} [\ln|u|]_5^{17}\).

Therefore, the evaluation is:

\(\frac{1}{2} (\ln|17| - \ln|5|)\).

By the properties of logarithms, this simplifies to:

\(\frac{1}{2} \ln \left(\frac{17}{5}\right)\).

So, the value of the integral is \(\frac{1}{2} \ln \left(\frac{17}{5}\right)\), which matches the correct answer \(\frac{1}{2} \log\left(\frac{17}{5}\right)\).