Not as easy as it seems

Calculus Level 3

Calculate the following definite integral: ∫ 0 4 t 4 1 + t 4 d t \large \int_0^4\frac{t^4}{1+t^4}dt Hint: Use partial fractions.

Caution: This is a long and tedious integration. Be patient and good luck!


The answer is 2.8945.

Muhammad Arifur Rahman by Muhammad Arifur Rahman , 23, Bangladesh

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1 solution

Chew-Seong Cheong
Dec 18, 2018

I = ∫ 0 4 t 4 1 + t 4 d t = ∫ 0 4 ( 1 − 1 1 + t 4 ) d t = 4 − ∫ 0 4 1 1 + t 4 d t = 4 − ∫ 0 4 1 ( t 2 − 2 t + 1 ) ( t 2 + 2 t + 1 ) d t = 4 − 1 2 2 ∫ 0 4 ( t + 2 t 2 + 2 t + 1 − t − 2 t 2 − 2 t + 1 ) d t = 4 − 1 4 2 ∫ 0 4 ( 2 t + 2 + 2 t 2 + 2 t + 1 − 2 t − 2 − 2 t 2 − 2 t + 1 ) d t = 4 − 1 4 2 ∫ 0 4 ( 2 t + 2 t 2 + 2 t + 1 − 2 t − 2 t 2 − 2 t + 1 ) d t − 1 4 ∫ 0 4 ( 1 t 2 + 2 t + 1 + 1 t 2 − 2 t + 1 ) d t = 4 − 1 4 2 ln ⁡ ( t 2 + 2 t + 1 t 2 − 2 t + 1 ) ∣ 0 4 − 1 2 ∫ 0 4 ( 1 ( 2 t + 1 ) 2 + 1 + 1 ( 2 t − 1 ) 2 + 1 ) d t = 4 − 1 4 2 ln ⁡ ( 17 + 4 2 17 − 4 2 ) − 1 2 2 ( tan ⁡ − 1 ( 2 t + 1 ) + tan ⁡ − 1 ( 2 t − 1 ) ) ∣ 0 4 = 4 − 1 4 2 ln ⁡ ( 17 + 4 2 17 − 4 2 ) − 1 2 2 ( tan ⁡ − 1 ( 4 2 + 1 ) + tan ⁡ − 1 ( 4 2 − 1 ) ) ≈ 2.895 \begin{aligned} I & = \int_0^4 \frac {t^4}{1+t^4}dt \\ & = \int_0^4 \left(1 - \frac 1{1+t^4} \right) dt \\ & = 4 - \int_0^4 \frac 1{1+t^4} dt \\ & = 4 - \int_0^4 \frac 1{(t^2-\sqrt 2t +1)(t^2+\sqrt 2t+1)}dt \\ & = 4 - \frac 1{2\sqrt 2} \int_0^4 \left(\frac {t+\sqrt 2}{t^2+\sqrt 2t +1} - \frac {t-\sqrt 2}{t^2-\sqrt 2t +1} \right) dt \\ & = 4 - \frac 1{4\sqrt 2} \int_0^4 \left(\frac {2t+\sqrt 2+\sqrt 2}{t^2+\sqrt 2t +1} - \frac {2t-\sqrt 2-\sqrt 2}{t^2-\sqrt 2t +1} \right) dt \\ & = 4 - \frac 1{4\sqrt 2} \int_0^4 \left(\frac {2t+\sqrt 2}{t^2+\sqrt 2t +1} - \frac {2t-\sqrt 2}{t^2-\sqrt 2t +1} \right) dt - \frac 14 \int_0^4 \left(\frac 1{t^2+\sqrt 2t +1} + \frac 1{t^2-\sqrt 2t +1} \right) dt \\ & = 4 - \frac 1{4\sqrt 2} \ln \left(\frac {t^2+\sqrt 2t+1}{t^2-\sqrt 2t + 1}\right)\bigg|_0^4 - \frac 12 \int_0^4 \left(\frac 1{(\sqrt 2t +1)^2+1} +\frac 1{(\sqrt 2t-1)^2+1} \right) dt \\ & = 4 - \frac 1{4\sqrt 2} \ln \left(\frac {17+4\sqrt 2}{17-4\sqrt 2} \right) - \frac 1{2\sqrt 2} \left(\tan^{-1} (\sqrt 2t+1) + \tan^{-1} (\sqrt2t-1)\right) \bigg|_0^4 \\ & = 4 - \frac 1{4\sqrt 2} \ln \left(\frac {17+4\sqrt 2}{17-4\sqrt 2} \right) - \frac 1{2\sqrt 2} \left(\tan^{-1} (4\sqrt 2+1) + \tan^{-1} (4\sqrt2-1) \right) \\ & \approx \boxed{2.895} \end{aligned}

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Thanks for taking your time to write such a detailed solution, sir.

Muhammad Arifur Rahman - 2 years, 5 months ago

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You are welcome.

Chew-Seong Cheong - 2 years, 5 months ago

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