Computing Indefinite Integrals — Question 4

Question 4 - Solution

Use substitution. Let $$u = x^2+4, \qquad du = 2x\,dx.$$ Then $$\int x\sqrt{x^2+4}\,\ln(x^2+4)\,dx = \frac12\int u^{1/2}\ln u\,du.$$

Apply integration by parts with $$w=\ln u, \qquad dv=u^{1/2}\,du.$$ Then $$dw=\frac{1}{u}\,du, \qquad v=\frac{2}{3}u^{3/2}.$$

Thus, $$\int u^{1/2}\ln u\,du = \frac{2}{3}u^{3/2}\ln u - \frac{2}{3}\int u^{1/2}\,du = \frac{2}{3}u^{3/2}\ln u - \frac{4}{9}u^{3/2}.$$

Multiply by $\tfrac12$ and substitute back: $$\boxed{ \frac{1}{3}(x^2+4)^{3/2}\ln(x^2+4) -\frac{2}{9}(x^2+4)^{3/2} + C }$$