Question

Find the area in the first quadrant under the curve y = 1 / (x^2+6x+10)

Answer 1

The area is given by


`\displaystyle\int_0^\infty(\mathrm dx)/(x^2+6x+10)`

Complete the square in the denominator:


`x^2+6x+10=x^2+6x+9+1=(x+3)^2+1`

then substitute
`x+3=\tan y`, so that
`\mathrm dx=\sec^2y\,\mathrm dy`. Then as
`x\to0^+`, we have
`y\to\arctan3`; as
`x\to\infty`, we have
`y\to\frac\pi2`. So the integral becomes


`\displaystyle\int_(\arctan3)^(\pi/2)(\sec^2y)/(\tan^2y+1)\,\mathrm dy=\int_(\arctan3)^(\pi/2)\mathrm dy=\frac\pi2-\arctan3`