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

User Raimon
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1 Answer

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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
User Q The Platypus
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