What To Do To The Integral?

Calculus Level 3

Evaluate

∫ 0 ∞ ln ⁡ x x 2 + x + 1 d x \int_0^\infty \frac{ \ln x } { x^2 + x + 1 } \, dx


The answer is 0.00.

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Calvin Lin by Calvin Lin

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2 solutions

Chew-Seong Cheong
Nov 14, 2016

Relevant wiki: Integration Tricks

I = ∫ 0 ∞ ln ⁡ x x 2 + x + 1 d x Using the identity ∫ 0 ∞ f ( x ) x 2 d x = ∫ 0 ∞ f ( x ) d x = 1 2 ∫ 0 ∞ ln ⁡ x x 2 + x + 1 d x + ln ⁡ ( 1 x ) x 2 ( 1 x 2 + 1 x + 1 ) d x = 1 2 ∫ 0 ∞ ln ⁡ x x 2 + x + 1 d x − ln ⁡ x 1 + x + x 2 d x = 0 \begin{aligned} I & = \int_0^\infty \frac {\ln x}{x^2+x+1} dx & \small {\color{#3D99F6}\text{Using the identity }\int_0^\infty \frac {f(x)}{x^2} dx = \int_0^\infty f(x) \ dx} \\ & = \frac 12 \int_0^\infty \frac {\ln x}{x^2+x+1} dx + \frac {\ln \left(\frac 1x\right)}{x^2 \left(\frac 1{x^2}+\frac 1x+1\right)} dx \\ & = \frac 12 \int_0^\infty \frac {\ln x}{x^2+x+1} dx - \frac {\ln x}{1+x+x^2} dx \\ & = \boxed{0} \end{aligned}

3 Helpful 0 Interesting 1 Brilliant 0 Confused
Prakhar Bindal
Nov 14, 2016

Put x = 1/t in the integral

You will get I = -I which implies I = 0

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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