More Substitution Rule — Question 9

Question 9

Evaluate the integral $$\int \frac{x+1}{(x^2+2x+3)^{3/2}}\,dx.$$

Question 9 - Solution

Begin by completing the square in the denominator: $$x^2+2x+3=(x+1)^2+2.$$

Let $$u=x+1.$$ Then $$du=dx, \qquad (x^2+2x+3)^{3/2}=(u^2+2)^{3/2}.$$

Substitute into the integral: $$\int \frac{u}{(u^2+2)^{3/2}}\,du.$$

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

Then $$\int \frac{u}{(u^2+2)^{3/2}}\,du = \frac12\int w^{-3/2}\,dw.$$

Integrate: $$\frac12\int w^{-3/2}\,dw = \frac12\left(-2w^{-1/2}\right) = -\frac{1}{\sqrt{w}}.$$

Substitute back: $$\boxed{ -\frac{1}{\sqrt{x^2+2x+3}} + C }$$