More Substitution Rule — Question 7

Question 7 - Solution

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

Let $$u=x+\tfrac12.$$ Then $$du=dx, \qquad \sqrt{x^2+x}=\sqrt{u^2-\tfrac14}.$$

The integral becomes $$\int \sqrt{u^2-\tfrac14}\,du.$$

Now use a standard substitution for expressions of the form $\sqrt{u^2-a^2}$. Let $$u=\tfrac12\sec\theta.$$ Then $$du=\tfrac12\sec\theta\tan\theta\,d\theta, \qquad \sqrt{u^2-\tfrac14}=\tfrac12\tan\theta.$$

Substitute: $$\int \sqrt{u^2-\tfrac14}\,du = \int \tfrac12\tan\theta\cdot\tfrac12\sec\theta\tan\theta\,d\theta = \frac14\int \tan^2\theta\sec\theta\,d\theta.$$

Use the identity $\tan^2\theta=\sec^2\theta-1$: $$\frac14\int (\sec^3\theta-\sec\theta)\,d\theta.$$

Integrate: $$\int \sec^3\theta\,d\theta =\frac12(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|),$$ $$\int \sec\theta\,d\theta =\ln|\sec\theta+\tan\theta|.$$

Thus, $$\frac14\left[ \frac12\sec\theta\tan\theta -\frac12\ln|\sec\theta+\tan\theta| \right].$$

Return to $x$ using $$\sec\theta=2u=2x+1, \qquad \tan\theta=2\sqrt{x^2+x}.$$

Final Answer: $$\boxed{ \frac12\left(x+\tfrac12\right)\sqrt{x^2+x} -\frac18\ln\!\left|2x+1+2\sqrt{x^2+x}\right| + C }$$