A trickier antiderivative

Calculus Level 2

Evaluate

∫ − ∞ ∞ d x x 2 + 1 \int_{-\infty}^\infty \frac {dx}{x^2 + 1}

Round answer to 2 decimal places.


The answer is 3.14.

Denton Young by Denton Young , 47, USA

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2 solutions

Chew-Seong Cheong
Sep 17, 2019

I = ∫ − ∞ ∞ d x x 2 + 1 Since the integrand is even = 2 ∫ 0 ∞ d x x 2 + 1 = 2 tan ⁡ − 1 x ∣ 0 ∞ = π ≈ 3.14 \begin{aligned} I & = \int_{-\infty}^\infty \frac {dx}{x^2+1} & \small \color{#3D99F6} \text{Since the integrand is even} \\ & = 2 \int_0^\infty \frac {dx}{x^2+1} \\ & = 2 \tan^{-1} x \ \big|_0^\infty \\ & = \pi \approx \boxed{3.14} \end{aligned}

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Denton Young
Sep 16, 2019

The antiderivative of the given function is arctan x.

So the answer is π / 2 \pi/2 - ( − π / 2 ) (-\pi/2) = π \pi

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If anyone is confused on why the anti-derivative is arctan(x), use trig substitution

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