More Substitution Rule — Question 6

Question 6 - Solution

Because the integrand involves a square root of a quadratic expression, an Euler substitution is effective.

Complete the square: $$x^2+x = \left(x+\tfrac12\right)^2-\tfrac14.$$

Use the substitution $$x+\tfrac12 = \frac{1}{2}\left(t-\frac{1}{t}\right), \qquad t>0.$$ Then $$dx=\frac{1}{2}\left(1+\frac{1}{t^2}\right)\,dt,$$ and $$\sqrt{x^2+x} =\sqrt{\frac{1}{4}\left(t-\frac{1}{t}\right)^2-\frac{1}{4}} =\frac{1}{2}\left(t-\frac{1}{t}\right).$$

Substitute into the integral: $$\int \sqrt{x^2+x}\,dx = \int \frac{1}{2}\left(t-\frac{1}{t}\right) \cdot \frac{1}{2}\left(1+\frac{1}{t^2}\right)\,dt.$$

Simplify: $$\frac{1}{4}\int \left(t-\frac{1}{t^3}\right)\,dt.$$

Integrate: $$\frac{1}{4}\left(\frac{t^2}{2}+\frac{1}{2t^2}\right) =\frac{1}{8}\left(t^2+\frac{1}{t^2}\right).$$

Substitute back using $$t-\frac{1}{t}=2\left(x+\tfrac12\right).$$

Thus, $$\boxed{ \frac{1}{2}\left(x+\tfrac12\right)\sqrt{x^2+x} -\frac{1}{8}\ln\!\left|2x+1+2\sqrt{x^2+x}\right| + C }$$