Indefinite Integrals — Question 4

Question 4 - Solution

We first rewrite the integrand using exponents: $$\sqrt{x}\,\ln x + \frac{1}{\sqrt{x}} = x^{1/2}\ln x + x^{-1/2}.$$

For the first term, apply integration by parts with $$u=\ln x, \qquad dv=x^{1/2}\,dx.$$ Then $$du=\frac{1}{x}\,dx, \qquad v=\frac{2}{3}x^{3/2}.$$ Thus, $$\int x^{1/2}\ln x\,dx = \frac{2}{3}x^{3/2}\ln x - \frac{2}{3}\int x^{1/2}\,dx = \frac{2}{3}x^{3/2}\ln x - \frac{4}{9}x^{3/2}.$$

For the second term, $$\int x^{-1/2}\,dx = 2x^{1/2}.$$

Combining results: $$\boxed{ \frac{2}{3}x^{3/2}\ln x - \frac{4}{9}x^{3/2} + 2x^{1/2} + C }$$