Show solutionHide solution+ Question 8 Evaluate the definite integral ∫012x(1+x2)2dx.\int_{0}^{1} \frac{2x}{\left(1+x^2\right)^2}\,dx. Question 8 - Solution The integrand suggests a substitution involving 1+x21+x^2. Let u=1+x2.u=1+x^2. Then du=2xdx.du=2x\,dx. Change the limits of integration. When x=0x=0, u=1u=1. When x=1x=1, u=2u=2. Substitute into the integral: ∫012x(1+x2)2dx=∫12u−2du.\int_{0}^{1} \frac{2x}{(1+x^2)^2}\,dx = \int_{1}^{2} u^{-2}\,du. Integrate: ∫u−2du=−u−1.\int u^{-2}\,du = -\,u^{-1}. Apply the limits: −u−1|12=−12−(−1)=12.-\,u^{-1}\Big|_{1}^{2} = -\frac{1}{2}-(-1) = \frac{1}{2}. 12\boxed{\frac{1}{2}}