# Thread: Challenge Integral

1. ## Challenge Integral

Calculate $\displaystyle \int_{0}^{1}\frac{x^4+1}{2 x^4-4 x^3+6 x^2-4 x+3}\;{dx}.$

2. Hint:

Spoiler:
\displaystyle \begin{aligned} 2 x^4-4 x^3+6 x^2-4 x+3} & = x^4+(x^4-4x^3+6x^2-4x+1)+2 \\& = x^4+(1-x)^4+2. \end{aligned}

3. Okay, following from the hint, here is the solution:

\displaystyle \displaystyle \begin{aligned} I_{n} & = \int_{0}^{a}\frac{x^n+b}{x^n+(a-x)^n+2b}\;{dx} \\& = \int_{0}^{a}\frac{(a-x)^n+b}{(a-x)^n+x^n+2b}\;{dx} \\& = \frac{1}{2}\int_{0}^{a}\frac{x^n+(a-x)^n+2b}{x^n+(a-x)^n+2b}\;{dx} \\& = \frac{1}{2}~a.\end{aligned}

In general, we have the following nice result:

\displaystyle \displaystyle \begin{aligned} J & = \int_{0}^{a} \frac{f(x)}{f(x)+f(a-x)}\;{dx} \\& = \int_{0}^{a}\frac{f(a-x)}{f(a-x)+f(x)}\;{dx} \\&= \frac{1}{2}\int_{0}^{a} \frac{f(x)+f(a-x)}{f(x)+f(a-x)}\;{dx} \\& = \frac{1}{2}~a. \end{aligned}.

4. It's funny how you phrase the title, so methinks that it's actually an algebra problem rather than a calculus problem haha, and of course, having in mind the old symmetry trick.

5. Originally Posted by TheCoffeeMachine
Okay, following from the hint, here is the solution:

\displaystyle \displaystyle \begin{aligned} I_{n} & = \int_{0}^{a}\frac{x^n+b}{x^n+(a-x)^n+2b}\;{dx} \\& = \int_{0}^{a}\frac{(a-x)^n+b}{(a-x)^n+x^n+2b}\;{dx} \\& = \frac{1}{2}\int_{0}^{a}\frac{x^n+(a-x)^n+2b}{x^n+(a-x)^n+2b}\;{dx} \\& = \frac{1}{2}~a.\end{aligned}

In general, we have the following nice result:

\displaystyle \displaystyle \begin{aligned} J & = \int_{0}^{a} \frac{f(x)}{f(x)+f(a-x)}\;{dx} \\& = \int_{0}^{a}\frac{f(a-x)}{f(a-x)+f(x)}\;{dx} \\&= \frac{1}{2}\int_{0}^{a} \frac{f(x)+f(a-x)}{f(x)+f(a-x)}\;{dx} \\& = \frac{1}{2}~a. \end{aligned}.
Dear TheCoffeeMachine,

Thanks for posting this challenge integral. I enjoyed solving it. Alternatively we could also get the answer if we use use the substitution, $\displaystyle x=1-y$

$\displaystyle \int_{0}^{1}\frac{x^4+1}{2 x^4-4 x^3+6 x^2-4 x+3}\;{dx}$

$\displaystyle =-\int_{1}^{0}\frac{(y-1)^4+1}{y^4+1+(y-1)^4+1}dy$

$\displaystyle =\int_{0}^{1}\frac{(y-1)^4+1}{y^4+1+(y-1)^4+1}dy$

Hence, $\displaystyle \int_{0}^{1}\frac{x^4+1}{x^4+1+(x-1)^4+1}dx=\int_{0}^{1}\frac{(x-1)^4+1}{x^4+1+(x-1)^4+1}dx$-----------(1)

$\displaystyle \int_{0}^{1}\frac{x^4+1}{x^4+1+(x-1)^4+1}dx+\int_{0}^{1}\frac{(x-1)^4+1}{x^4+1+(x-1)^4+1}dx=1$----------(2)

By (1) and (2),

$\displaystyle \int_{0}^{1}\frac{x^4+1}{2 x^4-4 x^3+6 x^2-4 x+3}\;{dx}=0.5$