More Substitution Rule — Question 8

Question 8 - Solution

Complete the square in the denominator: $$x^2+4x+8=(x+2)^2+4.$$

Let $$u=x+2.$$ Then $$du=dx, \qquad x+2=u.$$

Substitute into the integral: $$\int \frac{x+2}{\sqrt{x^2+4x+8}}\,dx = \int \frac{u}{\sqrt{u^2+4}}\,du.$$

Now use a direct substitution. Let $$w=u^2+4, \qquad dw=2u\,du.$$

Then $$\int \frac{u}{\sqrt{u^2+4}}\,du = \frac12\int w^{-1/2}\,dw = \sqrt{w}.$$

Substitute back: $$\boxed{ \sqrt{x^2+4x+8} + C }$$