Computing Definite Integrals — Question 10

Question 10 - Solution

This integral suggests a substitution because the derivative of the denominator appears in the numerator.

Let $$u=1+x^2.$$ Then $$du=2x\,dx.$$

Change the limits of integration. When $x=0$, $u=1$. When $x=1$, $u=2$.

Substitute into the integral: $$\int_{0}^{1} \frac{2x}{1+x^2}\,dx = \int_{1}^{2} \frac{1}{u}\,du.$$

Integrate: $$\int \frac{1}{u}\,du=\ln u.$$

Apply the limits: $$\ln u\Big|_{1}^{2}=\ln 2.$$

$$\boxed{\ln 2}$$