Question 1 Evaluate the definite integral ∫01x1+x2dx.\int_{0}^{1} \frac{x}{\sqrt{1+x^2}}\,dx. Show solutionHide solution+Question 1 - Solution Use substitution to simplify the integrand. Let u=1+x2.u=1+x^2. Then du=2xdx⇒xdx=12du.du=2x\,dx \quad\Rightarrow\quad x\,dx=\frac12\,du. Change the limits of integration. When x=0x=0, u=1u=1. When x=1x=1, u=2u=2. Substitute into the integral: ∫01x1+x2dx=12∫12u−1/2du.\int_{0}^{1} \frac{x}{\sqrt{1+x^2}}\,dx = \frac12\int_{1}^{2} u^{-1/2}\,du. Integrate: 12∫u−1/2du=12⋅2u1/2=u.\frac12\int u^{-1/2}\,du = \frac12\cdot 2u^{1/2} = \sqrt{u}. Apply the new limits: u|12=2−1.\sqrt{u}\Big|_{1}^{2} = \sqrt{2}-1. 2−1\boxed{\sqrt{2}-1}