Answer: ln5 - ln2 ≈ 0.919
Step-by-step explanation:
[tex]\int\limits^2_1 {\dfrac{2x}{x^2+1}} \, dx\qquad \qquad =\int\limits^2_1 {(x^2+1)^{-1}} \, 2xdx[/tex]
Let u = x² + 1 → du = 2x
[tex]\text{Then we have:}\quad \int\limits^2_1 {u^{-1}} \, du \qquad =ln|u|\bigg|^2_1[/tex]
Substitute u = x² + 1
[tex]ln|x^2+1|\bigg|^2_1\quad =ln|2^2+1|-ln|1^2+1|\quad = ln|5|-ln|2|[/tex]
