Computing Indefinite Integrals — Question 10

Question 10

Evaluate the indefinite integral: $$\int \frac{x^2+2x+2}{\sqrt{x^2+2x+5}}\,dx.$$

Question 10 - Solution

We are given: $$\int \frac{x^2+2x+2}{\sqrt{x^2+2x+5}}\,dx$$

Rewrite the numerator: $$x^2+2x+2=(x^2+2x+5)-3.$$

Then $$\int \frac{x^2+2x+2}{\sqrt{x^2+2x+5}}\,dx = \int \sqrt{x^2+2x+5}\,dx - 3\int \frac{dx}{\sqrt{x^2+2x+5}}.$$

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

Let $$t=x+1.$$

Then $$x^2+2x+5=t^2+4$$ and $$dt=dx.$$

So the integral becomes: $$\int \sqrt{t^2+4}\,dt - 3\int \frac{dt}{\sqrt{t^2+4}}.$$

Using the formulas $$\int \sqrt{t^2+a^2}\,dt = \frac{t}{2}\sqrt{t^2+a^2} + \frac{a^2}{2} \ln\left|t+\sqrt{t^2+a^2}\right|$$ and $$\int \frac{dt}{\sqrt{t^2+a^2}} = \ln\left|t+\sqrt{t^2+a^2}\right|,$$ with $a=2$, we get: $$\int \sqrt{t^2+4}\,dt = \frac{t}{2}\sqrt{t^2+4} + 2\ln\left|t+\sqrt{t^2+4}\right|$$ and $$3\int \frac{dt}{\sqrt{t^2+4}} = 3\ln\left|t+\sqrt{t^2+4}\right|.$$

Therefore, $$\int \frac{x^2+2x+2}{\sqrt{x^2+2x+5}}\,dx = \frac{t}{2}\sqrt{t^2+4} + 2\ln\left|t+\sqrt{t^2+4}\right| - 3\ln\left|t+\sqrt{t^2+4}\right|.$$

Simplify: $$= \frac{t}{2}\sqrt{t^2+4} - \ln\left|t+\sqrt{t^2+4}\right| + C.$$

Substitute back $t=x+1$: $$\boxed{ \frac{x+1}{2}\sqrt{x^2+2x+5} - \ln\left|x+1+\sqrt{x^2+2x+5}\right| + C }$$