Show solutionHide solution+ Question 4 Evaluate the integral ∫xx2+4ln(x2+4)dx.\int x\,\sqrt{x^2+4}\,\ln(x^2+4)\,dx. Question 4 - Solution Use substitution. Let u=x2+4,du=2xdx.u = x^2+4, \qquad du = 2x\,dx. Then ∫xx2+4ln(x2+4)dx=12∫u1/2lnudu.\int x\sqrt{x^2+4}\,\ln(x^2+4)\,dx = \frac12\int u^{1/2}\ln u\,du. Apply integration by parts with w=lnu,dv=u1/2du.w=\ln u, \qquad dv=u^{1/2}\,du. Then dw=1udu,v=23u3/2.dw=\frac{1}{u}\,du, \qquad v=\frac{2}{3}u^{3/2}. Thus, ∫u1/2lnudu=23u3/2lnu−23∫u1/2du=23u3/2lnu−49u3/2.\int u^{1/2}\ln u\,du = \frac{2}{3}u^{3/2}\ln u - \frac{2}{3}\int u^{1/2}\,du = \frac{2}{3}u^{3/2}\ln u - \frac{4}{9}u^{3/2}. Multiply by 12\tfrac12 and substitute back: 13(x2+4)3/2ln(x2+4)−29(x2+4)3/2+C\boxed{ \frac{1}{3}(x^2+4)^{3/2}\ln(x^2+4) -\frac{2}{9}(x^2+4)^{3/2} + C }