# Problem of the Week

## Updated at Dec 2, 2013 12:09 PM

To get more practice in calculus, we brought you this problem of the week:

How can we find the integral of $$\frac{4x+8}{2{x}^{2}+8x+3}$$?

Check out the solution below!

$\int \frac{4x+8}{2{x}^{2}+8x+3} \, dx$

 1 Use Integration by Substitution.Let $$u=2{x}^{2}+8x+3$$, $$du=4x+8 \, dx$$2 Using $$u$$ and $$du$$ above, rewrite $$\int \frac{4x+8}{2{x}^{2}+8x+3} \, dx$$.$\int \frac{1}{u} \, du$3 The derivative of $$\ln{x}$$ is $$\frac{1}{x}$$.$\ln{u}$4 Substitute $$u=2{x}^{2}+8x+3$$ back into the original integral.$\ln{(2{x}^{2}+8x+3)}$5 Add constant.$\ln{(2{x}^{2}+8x+3)}+C$Doneln(2*x^2+8*x+3)+C