Computing Definite Integrals — Question 4

Question 4 - Solution

This integral is well-suited for a direct substitution because the derivative of $1+x^2$ appears in the numerator.

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 $x=1$, $u=2$.

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

Integrate: $$\frac12\int u^{-1/2}\,du = \frac12\cdot 2u^{1/2} = \sqrt{u}.$$

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

$$\boxed{\sqrt{2}-1}$$