Substitution Rule for Definite Integrals — Question 6

Question 6 - Solution

The integrand suggests a substitution involving $1+x^2$.

Let $$u=1+x^2.$$ Then $$du=2x\,dx \quad\Rightarrow\quad x\,dx=\frac12\,du.$$

Change the limits of integration. When $x=0$, $u=1$. When $x=\sqrt{3}$, $u=4$.

Substitute into the integral: $$\int_{0}^{\sqrt{3}} \frac{x}{1+x^2}\,dx = \frac12\int_{1}^{4} \frac{1}{u}\,du.$$

Integrate: $$\frac12\int \frac{1}{u}\,du = \frac12\ln u.$$

Apply the limits: $$\frac12\ln u\Big|_{1}^{4} = \frac12\ln 4 = \ln 2.$$

$$\boxed{\ln 2}$$