Question

I have NO IDEA how to do! Please help with this one!!!! 50 pts

Answer 1

`\displaystyle \int (dx)/(5x^2 -10x +15)\\\\\\=\displaystyle \int (dx)/(5(x^2-2x+3))\\\\\\=\frac 1 5 \displaystyle \int (dx)/(x^2-2x+3)\\\\\\=\frac 15 \displaystyle \int (dx)/(x^2 -2x +1^2 -1 +3)\\\\\\=\frac 15 \displaystyle \int (dx)/((x-1)^2+2)\\\\\\=\frac 15 \displaystyle \int (du)/(u^2 +\left(\sqrt 2 \right)^2)~~~~~~~~~;[x-1 =u \implies dx = du]\\\\\\\\`

`=\frac 1 {5\sqrt 2} \tan^(-1) \left( \frac u{\sqrt 2}\right)+C~~~~~~;\left[\displaystyle \int (dx)/(a^2 +x^2) = \frac 1a \tan^(-1) \left( \frac xa \right) +C\right]\\\\\\=(1)/(5\sqrt 2) \tan^(-1) \left( (x-1)/(\sqrt 2)\right)+C`