Look familiar? (My seventeenth integral problem)

Calculus Level 5

$\large \displaystyle \int_{0}^{1} \dfrac{1-x^{3}}{1+x^{3}} \, dx$

If the above integral can be expressed in the form

$-a + \dfrac{b \pi \sqrt{c}}{d} + \dfrac{f\ln 2}{g}$

where $a, b, c, d, f, g$ are positive integers and $\gcd(b, d)=\gcd(f, g)= 1$ , find $a+b+c+d+f+g$ .

The answer is 20.

1 solution

Rishabh Jain
Feb 27, 2016

$\mathfrak{I}=\int_0^1(\dfrac{1+x^3}{1-x^3})dx=\int_0^1(-1+\dfrac{2}{1+x^3})dx$ $\mathfrak{S}=\int_0^1 (\dfrac{1}{1+x^3})dx$ $=\dfrac{1}{3}\int_0^1(\dfrac{1}{1+x}+\frac{1}{2}(\dfrac{-2x+1}{x^2-x+1}+\dfrac{3}{x^2-x+1}))dx$ See this for Integration using Partial fractions $=\dfrac 13(\ln(1+x)-\frac 12(\ln (x^2-x+1)+2\sqrt 3(\tan^{-1}(\dfrac{2x-1}{\sqrt 3})))|_0^1$ $=\dfrac 13 \ln 2+\dfrac{\pi\sqrt 3}{9}$ Hence, $\mathfrak I=-1+2\mathfrak S$ $\large =-1+\dfrac 23 \ln 2+\dfrac{2\pi\sqrt 3}{9}$ $\therefore 1+2+3+9+2+3=\huge\boxed{20}$

how to simplify $\dfrac{3}{x^2-x+1}\dx$ ?

Anik Mandal - 5 years, 3 months ago

$\dfrac{3}{x^2-x+1}=\dfrac{3}{(x-\frac 12)^2+\frac 34}$ And then use : $\int \dfrac{dx}{x^2+a^2}=\dfrac 1a (\tan^{-1}\frac xa)$ and apply appropriate limits.

Rishabh Jain - 5 years, 3 months ago

Thanks for the explanation

Anik Mandal - 5 years, 3 months ago

Look familiar? (My seventeenth integral problem)

The answer is 20.

1 solution

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