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Kindly help me with this

evaluate the integral

\gamma \binom{2}{1} \frac{2x}{{x}^(2) + 1} dx


1 Answer

4 votes

Answer: ln5 - ln2 ≈ 0.919

Explanation:


\int\limits^2_1 {(2x)/(x^2+1)} \, dx\qquad \qquad =\int\limits^2_1 {(x^2+1)^(-1)} \, 2xdx

Let u = x² + 1 → du = 2x


\text{Then we have:}\quad \int\limits^2_1 {u^(-1)} \, du \qquad =ln|u|\bigg|^2_1

Substitute u = x² + 1


ln|x^2+1|\bigg|^2_1\quad =ln|2^2+1|-ln|1^2+1|\quad = ln|5|-ln|2|

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