Question

Complete the square to evaluate the definite integral
∫dx/x^2-2x+5

Bounds:1-3

Answer 1

`\displaystyle \int_1^3 (\mathrm dx)/(x^2-2x+5)`

Follow the instruction and complete the square in the denominator:

x ² - 2x + 5 = (x ² - 2x + 1) + 4 = (x - 1)² + 4

Then the integral is

`\displaystyle \int_(x=1)^(x=3) (\mathrm dx)/((x-1)^2+4)`

Substitute y = x - 1 and dy = dx :

`\displaystyle \int_(y+1=1)^(y+1=3) (\mathrm dy)/(y^2+4) = \int_(y=0)^(y=2)(\mathrm dy)/(y^2+4)`

Substitute y = 2 tan(z) and dy = 2 sec²(z) dz :

`\displaystyle \int_(2\tan(z)=0)^(2\tan(z)=2)(2\sec^2(z))/((2\tan(z))^2+4)\,\mathrm dz = \frac12 \int_(z=0)^(z=\pi/4) (\sec^2(z))/(\tan^2(z)+1)\,\mathrm dz \\\\ = \frac12 \int_(z=0)^(z=\pi/4) (\sec^2(z))/(\sec^2(z))\,\mathrm dz \\\\ = \frac12 \int_(z=0)^(z=\pi/4) \mathrm dz = \frac12 z\bigg|_(z=0)^(z=\pi/4) = \frac12 \left(\frac\pi4-0\right) = \boxed{\frac\pi8}`