Show solutionHide solution+ Question 5 Evaluate the integral ∫2x1+1+x2dx.\int \frac{2x}{1+\sqrt{1+x^2}}\,dx. Question 5 - Solution Use substitution. Let u=1+x2.u=\sqrt{1+x^2}. Then u2=1+x2⇒2udu=2xdx⇒2xdx=2udu.u^2=1+x^2 \quad\Rightarrow\quad 2u\,du=2x\,dx \quad\Rightarrow\quad 2x\,dx=2u\,du. Substitute into the integral: ∫2x1+1+x2dx=∫2u1+udu.\int \frac{2x}{1+\sqrt{1+x^2}}\,dx = \int \frac{2u}{1+u}\,du. Rewrite the integrand: 2u1+u=2(1−11+u).\frac{2u}{1+u} = 2\left(1-\frac{1}{1+u}\right). Integrate: ∫2(1−11+u)du=2u−2ln(1+u).\int 2\left(1-\frac{1}{1+u}\right)\,du = 2u-2\ln(1+u). Substitute back: 21+x2−2ln(1+1+x2)+C\boxed{ 2\sqrt{1+x^2} -2\ln\!\left(1+\sqrt{1+x^2}\right) + C }