Indefinite Integrals — Question 7

Question 7 - Solution

Combine the terms into a single fraction: $$\frac{\sqrt{1+2x}}{x} - \frac{1}{x\sqrt{1+2x}} = \frac{(1+2x)-1}{x\sqrt{1+2x}} = \frac{2x}{x\sqrt{1+2x}} = \frac{2}{\sqrt{1+2x}}.$$

Thus the integral simplifies to $$\int \frac{2}{\sqrt{1+2x}}\,dx.$$

Let $$u=1+2x, \qquad du=2\,dx.$$ Then $$\int \frac{2}{\sqrt{1+2x}}\,dx = \int u^{-1/2}\,du = 2u^{1/2}.$$

Substitute back: $$\boxed{ 2\sqrt{1+2x}+C }$$