More Substitution Rule — Question 4

Question 4 - Solution

Complete the square: $$x^2+2x+5=(x+1)^2+4.$$

Let $$u=x+1, \qquad du=dx, \qquad x=u-1.$$

Then the integral becomes $$\int \frac{u-1}{\sqrt{u^2+4}}\,du = \int \frac{u}{\sqrt{u^2+4}}\,du - \int \frac{1}{\sqrt{u^2+4}}\,du.$$

First integral: Let $w=u^2+4$, so $dw=2u\,du$: $$\int \frac{u}{\sqrt{u^2+4}}\,du = \frac12\int w^{-1/2}\,dw = \sqrt{u^2+4}.$$

Second integral (expanded): $$\int \frac{1}{\sqrt{u^2+4}}\,du.$$

Use a trigonometric substitution. Let $$u=2\tan\theta.$$ Then $$du=2\sec^2\theta\,d\theta, \qquad \sqrt{u^2+4}=\sqrt{4\tan^2\theta+4}=2\sec\theta.$$

Substitute: $$\int \frac{1}{\sqrt{u^2+4}}\,du = \int \frac{2\sec^2\theta}{2\sec\theta}\,d\theta = \int \sec\theta\,d\theta.$$

Recall: $$\int \sec\theta\,d\theta = \ln|\sec\theta+\tan\theta|.$$

Return to $u$: $$\tan\theta=\frac{u}{2}, \qquad \sec\theta=\frac{\sqrt{u^2+4}}{2}.$$

Thus, $$\int \frac{1}{\sqrt{u^2+4}}\,du = \ln\!\left(u+\sqrt{u^2+4}\right).$$

Final Answer: $$\boxed{ \sqrt{x^2+2x+5} - \ln\!\left(x+1+\sqrt{x^2+2x+5}\right) + C }$$