Question

How do you integrate this $\int\limits_0^1 {\dfrac{{{x^4}{{\left( {1 - x} \right)}^4}}}{{1 + {x^2}}}dx}$ ?

Answer 1

Hint : Integration is the process of finding the antiderivative. The integration of $g'\left( x \right)$ with respect to dx is given by $\int {g'\left( x \right)dx = g\left( x \right) + C}$ , where C is the constant of integration and we can find the integral of the given equation by expanding the numerator terms and then integrate the terms directly.
Formula used:
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
${\left( {a - b} \right)^4} = {a^4} - 4{a^3}b + 6{a^2}{b^2} - 4a{b^3} + {b^4}$

Complete step-by-step answer :
Let us write the given equation:
$\int\limits_0^1 {\dfrac{{{x^4}{{\left( {1 - x} \right)}^4}}}{{1 + {x^2}}}dx}$
Let,
$I = \int\limits_0^1 {\dfrac{{{x^4}{{\left( {1 - x} \right)}^4}}}{{1 + {x^2}}}dx}$
Expand the numerator terms as:
$I = \int\limits_0^1 {\dfrac{{\left( {{x^4} - 1} \right){{\left( {1 - x} \right)}^4} + {{\left( {1 - x} \right)}^4}}}{{1 + {x^2}}}dx}$
Here, $\left( {{x^4} - 1} \right)$ is of the form, ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ , hence we get
= \int\limits_0^1 {\left\\{ {\dfrac{{\left( {1 + {x^2}} \right)\left( {{x^2} - 1} \right){{\left( {1 - x} \right)}^4}}}{{\left( {1 + {x^2}} \right)}} + \dfrac{{{{\left( {1 - x} \right)}^4}}}{{\left( {1 + {x^2}} \right)}}} \right\\}dx}
As there are common terms involved, hence we get:
= \int\limits_0^1 {\left\\{ {\left( {{x^2} - 1} \right){{\left( {1 - x} \right)}^4} + \dfrac{{{{\left( {1 - x} \right)}^4}}}{{\left( {1 + {x^2}} \right)}}} \right\\}dx}
We can see that, ${\left( {1 - x} \right)^4}$ is of the form ${\left( {a - b} \right)^2}$ in which here, we have ${\left( {{{\left( {1 - x} \right)}^2}} \right)^2}$ , hence lets us apply: ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ i.e.,
= \int\limits_0^1 {\left\\{ {\left( {{x^2} - 1} \right){{\left( {1 - x} \right)}^4} + \dfrac{{{{\left( {1 + {x^2} - 2x} \right)}^2}}}{{\left( {1 + {x^2}} \right)}}} \right\\}dx}
Simplify the terms, as:
= \int\limits_0^1 {\left\\{ {\left( {{x^2} - 1} \right){{\left( {1 - x} \right)}^4} + \dfrac{{{{\left( {1 + {x^2}} \right)}^2} + {{\left( {2x} \right)}^2} - 2 \cdot 2x\left( {1 + {x^2}} \right)}}{{\left( {1 + {x^2}} \right)}}} \right\\}dx}
Simplifying the terms, with respect to the common terms we get:
= \int\limits_0^1 {\left\\{ {\left( {{x^2} - 1} \right){{\left( {1 - x} \right)}^4} + \left( {1 + {x^2}} \right) - 4x + \dfrac{{4x}}{{\left( {1 + {x^2}} \right)}}} \right\\}} dx
We know that ${\left( {1 - x} \right)^4}$ is of the form, ${\left( {a - b} \right)^4} = {a^4} - 4{a^3}b + 6{a^2}{b^2} - 4a{b^3} + {b^4}$ , hence apply this formula and simplify the terms of ${\left( {1 - x} \right)^4}$ as:
= \int\limits_0^1 {\left\\{ {\left( {{x^2} - 1} \right)\left( {{{\left( 1 \right)}^4} - 4{{\left( 1 \right)}^3}x + 6{{\left( 1 \right)}^2}{{\left( x \right)}^2} - 4 \cdot 1 \cdot {{\left( x \right)}^3} + {{\left( x \right)}^4}} \right) + \left( {1 + {x^2} - 4x} \right) + \dfrac{{4x}}{{1 + {x^2}}}} \right\\}} dx
= \int\limits_0^1 {\left\\{ {\left( {{x^2} - 1} \right)\left( {1 - 4x + 6{x^2} - 4{x^3} + {x^4}} \right) + \left( {1 + {x^2} - 4x} \right) + \dfrac{{4x}}{{1 + {x^2}}}} \right\\}} dx
Hence, combining and simplifying the common terms, we get
= $\int\limits_0^1 {\left( {{x^6} - 4{x^5} + 5{x^4} - 4{x^2} + \dfrac{{4x}}{{1 + {x^2}}}} \right)dx}$
Now, let us apply the integration directly:
$= \int\limits_0^1 {{x^6}dx - 4\int\limits_0^1 {{x^5}} dx + 5\int\limits_0^1 {{x^4}} dx - 4\int\limits_0^1 {{x^2}dx} + 4\int\limits_0^1 {\dfrac{x}{{1 + {x^2}}}dx} }$
Now, let us find the derivatives of x as:
$= \left[ {\dfrac{{{x^7}}}{7} - \dfrac{{4{x^6}}}{6} + \dfrac{{5{x^5}}}{5} - \dfrac{{4{x^3}}}{3}} \right] _0^1 + \int\limits_0^1 {\dfrac{{4\left( {1 + {x^2}} \right) - 4}}{{1 + {x^2}}}dx}$
$= \left[ {\dfrac{{{x^7}}}{7} - \dfrac{{4{x^6}}}{6} + \dfrac{{5{x^5}}}{5} - \dfrac{{4{x^3}}}{3}} \right] _0^1 + 4\int\limits_0^1 {dx - 4\int\limits_0^1 {\dfrac{{dx}}{{1 + {x^2}}}} }$
Applying integration to the terms we get:
$I = \left[ {\dfrac{{{x^7}}}{7} - \dfrac{{4{x^6}}}{6} + \dfrac{{5{x^5}}}{5} - \dfrac{{4{x^3}}}{3} + 4x - 4{{\tan }^{ - 1}}x} \right] _0^1$
Simplifying the terms, we get
$I = \dfrac{1}{7} - \dfrac{4}{6} + \dfrac{5}{5} - \dfrac{4}{3} + 4 - 4\left( {\dfrac{\pi }{4} - 0} \right)$
Therefore, the integration of the given function is:
$I = \dfrac{{22}}{7} - \pi$
Hence,
$\int\limits_0^1 {\dfrac{{{x^4}{{\left( {1 - x} \right)}^4}}}{{1 + {x^2}}}dx}$ = $\dfrac{{22}}{7} - \pi$
So, the correct answer is “ $\dfrac{{22}}{7} - \pi$ ”.

Note : To find the integration of these terms, we need to know all the respective formulas to expand, as the integration of these standard integrands is easily found using a direct form of integration method. There are different integration methods that are used to find an integral of some function, which is easier to evaluate the original integral. Hence, based on the function given we can find the integration of the function i.e., by using the integration methods.