Integral Integral7

Calculus Level 3

∫ 0 1 log ⁡ ( 1 x 4 + 1 ) d x = π a + b sinh ⁡ − 1 ( 1 ) + log ⁡ ( c ) \large \int_0^1 \log \left(\frac{1}{x^4}+1\right) \, dx=\frac{\pi }{\sqrt{a}}+\sqrt{b} \sinh ^{-1}(1)+\log (c)

where a a , b b , and c c are positive integers. Submit a + b + c a+b+c .


The answer is 6.

Christopher Criscitiello by Christopher Criscitiello , 25, USA

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1 solution

Chew-Seong Cheong
Dec 24, 2017

I = ∫ 0 1 log ⁡ ( 1 x 4 + 1 ) d x = ∫ 0 1 log ⁡ ( x 4 + 1 x 4 ) d x = ∫ 0 1 ( log ⁡ ( x 4 + 1 ) − 4 log ⁡ x ) d x = ∫ 0 1 log ⁡ ( x 4 + 1 ) d x − 4 ∫ 0 1 log ⁡ x d x By integration by parts. = x log ⁡ ( x 4 + 1 ) ∣ 0 1 − ∫ 0 1 4 x 4 x 4 + 1 d x − 4 [ x log ⁡ x − x ] 0 1 = log ⁡ 2 − 4 ∫ 0 1 ( 1 − 1 x 4 + 1 ) d x + 4 = log ⁡ 2 − 4 + 4 ∫ 0 1 1 ( x 2 + 2 x + 1 ) ( x 2 − 2 x + 1 ) d x + 4 = log ⁡ 2 + 4 ∫ 0 1 1 4 ( 2 x + 2 x 2 + 2 x + 1 − 2 x − 2 x 2 − 2 x + 1 ) d x = log ⁡ 2 + 1 2 ∫ 0 1 ( 2 x + 2 + 2 x 2 + 2 x + 1 − 2 x − 2 − 2 x 2 − 2 x + 1 ) d x = log ⁡ 2 + 1 2 ln ⁡ ( x 2 + 2 x + 1 x 2 − 2 x + 1 ) ∣ 0 1 + ∫ 0 1 1 x 2 + 2 x + 1 d x + ∫ 0 1 1 x 2 − 2 x + 1 d x = log ⁡ 2 + 1 2 ln ⁡ ( 2 + 2 2 − 2 ) + 2 ∫ 0 1 1 2 ( x + 1 2 ) 2 + 1 d x + 2 ∫ 0 1 1 2 ( x − 1 2 ) 2 + 1 d x = log ⁡ 2 + 1 2 ln ⁡ ( 3 + 2 2 ) + 2 ∫ 1 2 + 1 1 x 2 + 1 d x + 2 ∫ − 1 2 − 1 1 x 2 + 1 d x = log ⁡ 2 + 1 2 ln ⁡ ( 1 + 2 ) 2 + 2 ( tan ⁡ − 1 ( 2 + 1 ) − tan ⁡ − 1 1 + tan ⁡ − 1 ( 2 − 1 ) − tan ⁡ − 1 ( − 1 ) ) = log ⁡ 2 + 2 ln ⁡ ( 1 + 2 ) + 2 ( tan ⁡ − 1 2 + 1 + 2 − 1 1 − ( 2 + 1 ) ( 2 − 1 ) − π 4 + π 4 ) Note that sinh ⁡ − 1 x = ln ⁡ ( x + x 2 + 1 ) = log ⁡ 2 + 2 sinh ⁡ − 1 ( 1 ) + 2 × π 2 = π 2 + 2 sinh ⁡ − 1 ( 1 ) + log ⁡ 2 \begin{aligned} I & = \int_0^1 \log \left(\frac 1{x^4}+1\right) dx \\ & = \int_0^1 \log \left(\frac {x^4+1}{x^4} \right) dx \\ & = \int_0^1 \left(\log (x^4+1) - 4\log x \right) dx \\ & = {\color{#3D99F6} \int_0^1 \log (x^4+1)\ dx} - 4 \int_0^1 \log x\ dx & \small \color{#3D99F6} \text{By integration by parts.} \\ & = {\color{#3D99F6} x\log(x^4+1)\bigg|_0^1 - \int_0^1 \frac {4x^4}{x^4+1}\ dx} - 4\bigg[x \log x - x\bigg]_0^1 \\ & = {\color{#3D99F6} \log 2 - 4\int_0^1 \left(1 - \frac 1{x^4+1}\right) dx} + 4 \\ & = \log 2 - 4 + 4 \int_0^1 \frac 1{(x^2+\sqrt 2x+1)(x^2-\sqrt 2x+1)} dx + 4 \\ & = \log 2 + 4 \int_0^1 \frac 14 \left(\frac {\sqrt 2x+2}{x^2+\sqrt 2x+1}-\frac {\sqrt 2x-2}{x^2-\sqrt 2x+1}\right) dx \\ & = \log 2 + \frac 1{\sqrt 2} \int_0^1 \left(\frac {2x+\sqrt 2 + \sqrt 2}{x^2+\sqrt 2x+1}-\frac {2x-\sqrt 2 - \sqrt 2}{x^2-\sqrt 2x+1}\right) dx \\ & = \log 2 + \frac 1{\sqrt 2} \ln \left(\frac {x^2+\sqrt 2x+1}{x^2-\sqrt 2x+1} \right)\bigg|_0^1 + \int_0^1 \frac 1{x^2+\sqrt 2x+1} dx + \int_0^1 \frac 1{x^2-\sqrt 2x+1} dx \\ & = \log 2 + \frac 1{\sqrt 2} \ln \left(\frac {2+\sqrt 2}{2-\sqrt 2} \right) + 2 \int_0^1 \frac 1{2\left(x+\frac 1{\sqrt 2}\right)^2+1} dx + 2 \int_0^1 \frac 1{2\left(x-\frac 1{\sqrt 2}\right)^2+1} dx \\ & = \log 2 + \frac 1{\sqrt 2} \ln \left(3+2\sqrt 2 \right) + \sqrt 2 \int_1^{\sqrt 2+1} \frac 1{x^2+1} dx + \sqrt 2 \int_{-1}^{\sqrt 2-1} \frac 1{x^2+1} dx \\ & = \log 2 + \frac 1{\sqrt 2} \ln \left(1+\sqrt 2 \right)^2 + \sqrt 2 \left({\color{#3D99F6}\tan^{-1} (\sqrt 2+1)} - \tan^{-1} 1 + {\color{#3D99F6}\tan^{-1} (\sqrt 2-1)} - \tan^{-1} (-1)\right) \\ & = \log 2 + \sqrt 2 {\color{#D61F06} \ln \left(1+\sqrt 2 \right)} + \sqrt 2 \left({\color{#3D99F6} \tan^{-1} \frac {\sqrt 2+1 + \sqrt 2-1}{1-(\sqrt 2+1)(\sqrt 2-1)}} - \frac \pi 4 + \frac \pi 4 \right) & \small \color{#D61F06} \text{Note that }\sinh^{-1} x = \ln \left(x+\sqrt{x^2+1}\right) \\ & = \log 2 + \sqrt 2 {\color{#D61F06}\sinh^{-1}(1)} + \sqrt 2 \times \frac \pi 2 \\ & = \frac \pi {\sqrt 2} + \sqrt 2 \sinh^{-1}(1) + \log 2 \end{aligned}

Therefore, a + b + c = 2 + 2 + 2 = 6 a+b+c = 2+2+2 = \boxed{6}

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