Evaluate $\displaystyle \int{\dfrac{x^3+x}{x^4-9}}\,dx$

Two polynomials form a rational expression in this indefinite integration problem. The indefinite integral of the rational expression should be calculated with respect to $x$.

$\displaystyle \int{\dfrac{x^{\displaystyle 3}+x}{x^{\displaystyle 4}-9}}\,dx$

It is a proper fraction in algebraic form and let’s learn how to find the integral of the given rational function.

Find the integration by sum of integrals

In the numerator, two expressions are connected by a plus sign in the given function in fraction form. Hence, the function can be split as the sum of two fractions.

$=\,\,\,$ $\displaystyle \int{\bigg(\dfrac{x^{\displaystyle 3}}{x^{\displaystyle 4}-9}+\dfrac{x}{x^{\displaystyle 4}-9}\bigg)}\,dx$

Use the sum rule of the integration to find the integral of sum of two functions by the sum of their integrals.

$=\,\,\,$ $\displaystyle \int{\dfrac{x^{\displaystyle 3}}{x^{\displaystyle 4}-9}}\,dx$ $+$ $\displaystyle \int{\dfrac{x}{x^{\displaystyle 4}-9}}\,dx$

Let us assume that the integral of both functions are denoted by $I_{1}$ and $I_{2}$ respectively.

$\therefore\,\,\,$ $\displaystyle \int{\dfrac{x^{\displaystyle 3}+x}{x^{\displaystyle 4}-9}}\,dx$ $\,=\,$ $I_{1}$ $+$ $I_{2}$

Calculate the integral of first term

Now, let’s focus on finding the integral of the function in the first term.

$I_1$ $\,=\,$ $\displaystyle \int{\dfrac{x^{\displaystyle 3}}{x^{\displaystyle 4}-9}}\,dx$

The exponent of the function in numerator is $3$ and the exponent of the function in the first term of the denominator is $4$. It indicates that the expression in the numerator can be obtained by differentiating the expression in the denominator.

$=\,\,\,$ $\displaystyle \int{\dfrac{1 \times x^{\displaystyle 3}}{x^{\displaystyle 4}-9}}\,dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{x^{\displaystyle 4}-9}}\times x^{\displaystyle 3} \times dx$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{x^{\displaystyle 4}-9}}\times x^{\displaystyle 3}dx$

So, take $u \,=\, x^{\displaystyle 4}-9$

Now, differentiate the equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(u)}$ $\,=\,$ $\dfrac{d}{dx}{(x^{\displaystyle 4}-9)}$

Use the subtraction rule of differentiation to find the derivative of difference of the terms by the difference of their derivatives.

$\implies$ $\dfrac{d}{dx}{(u)}$ $\,=\,$ $\dfrac{d}{dx}{(x^{\displaystyle 4})}$ $-$ $\dfrac{d}{dx}{(9)}$

Now, use the power rule and the derivative rule of a constant to find the derivatives of the functions on the right-hand side of the equation.

$\implies$ $\dfrac{du}{dx}$ $\,=\,$ $4 \times x^{\displaystyle 4-1}$ $-$ $0$

$\implies$ $\dfrac{du}{dx} \,=\, 4 \times x^{\displaystyle 3}$

$\implies$ $du \,=\, 4 \times x^{\displaystyle 3} \times dx$

$\implies$ $4 \times x^{\displaystyle 3} \times dx \,=\, du$

$\implies$ $x^{\displaystyle 3} \times dx \,=\, \dfrac{du}{4}$

$\,\,\,\therefore\,\,\,\,\,\,$ $x^{\displaystyle 3}dx \,=\, \dfrac{du}{4}$

It is time to convert the first term in terms of $x$ into the function in terms of $u$.

$\displaystyle \int{\dfrac{1}{x^{\displaystyle 4}-9}}\times x^{\displaystyle 3}dx$

Consider $u \,=\, x^{\displaystyle 4}-9$ and $x^{\displaystyle 3}dx \,=\, \dfrac{du}{4}$ to convert the first term in terms of $u$.

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{u}}\times \dfrac{du}{4}$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{u}}\times \dfrac{1 \times du}{4}$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{u}}\times \dfrac{1}{4} \times du$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{4} \times \dfrac{1}{u}}\times du$

Exclude the constant factor in the first term from integration as per the constant multiple rule of integrals.

$=\,\,\,$ $\dfrac{1}{4} \times \displaystyle \int{\dfrac{1}{u}}\times du$

$=\,\,\,$ $\dfrac{1}{4}\displaystyle \int{\dfrac{1}{u}}\,du$

Use the reciprocal rule of integration to find the integral of the multiplicative inverse of variable $u$ with respect to $u$.

$=\,\,\,$ $\dfrac{1}{4}\big(\log_{e}{|u|}+c_1\big)$

$\therefore\,\,\,$ $I_1$ $\,=\,$ $\dfrac{1}{4}\Big(\log_{e}{\big|x^4-9\big|}+c_1\Big)$

Calculate the integral in the second term

In the numerator, there is a variable with exponent $1$ but the exponent of the variable in the denominator is $4$. The numerator cannot be obtained by differentiating the denominator in this case.

$I_2$ $\,=\,$ $\displaystyle \int{\dfrac{x}{x^{\displaystyle 4}-9}}\,dx$

However, this problem can be rectified by expressing the $x$ raised to the power $4$ as the square of $x$ squared because the numerator can be obtained by the differentiation of the square of variable $x$.

$=\,\,\,$ $\displaystyle \int{\dfrac{1 \times x}{x^{\displaystyle 2 \times 2}-9}}\,dx$

Now, use the power rule of exponents to express the first term in the denominator of the second term.

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{\Big(x^{\displaystyle 2}\Big)^{\displaystyle 2}-9}}\times xdx$

Now, take $v \,=\, x^{\displaystyle 2}$

Differentiate the expressions on both sides of the equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(v)} \,=\, \dfrac{d}{dx}{(x^{\displaystyle 2})}$

Use the power rule of differentiation to find the derivative of $x$ squared.

$\implies$ $\dfrac{dv}{dx} \,=\, 2x^{\displaystyle 2-1}$

$\implies$ $\dfrac{dv}{dx} \,=\, 2x^{\displaystyle 1}$

$\implies$ $\dfrac{dv}{dx} \,=\, 2x$

$\implies$ $dv \,=\, 2xdx$

$\implies$ $2xdx \,=\, dv$

$\implies$ $2 \times xdx \,=\, dv$

$\,\,\,\therefore\,\,\,\,\,\,$ $xdx \,=\, \dfrac{dv}{2}$

It is time to express the second term in terms of $v$.

$\displaystyle \int{\dfrac{1}{\Big(x^{\displaystyle 2}\Big)^{\displaystyle 2}-9}}\times xdx$

We have assumed that $v \,=\, x^{\displaystyle 2}$ and derived that $xdx \,=\, \dfrac{dv}{2}$. So, substitute them in the second term.

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{(v)^2-9}}\times \dfrac{dv}{2}$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{v^2-9}}\times \dfrac{dv}{2}$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{v^2-9}}\times \dfrac{1 \times dv}{2}$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{v^2-9}}\times \dfrac{1}{2}\times dv$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{2}\times \dfrac{1}{v^2-9}}\times dv$

It is time to separate the constant factor from the integral operation by using the constant multiple rule of integration.

$=\,\,\,$ $\dfrac{1}{2}\times \displaystyle \int{\dfrac{1}{v^2-9}}dv$

$=\,\,\,$ $\dfrac{1}{2}\times \displaystyle \int{\dfrac{1}{v^2-3^2}}dv$

The indefinite integration can be calculated by the reciprocal integral rule of difference of the squares.

$=\,\,\,$ $\dfrac{1}{2}\times \dfrac{1}{2 \times 3} \times \Bigg(\log_{e}{\bigg|\dfrac{v-3}{v+3}\bigg|}+c_2\Bigg)$

$=\,\,\,$ $\dfrac{1}{2}\times \dfrac{1}{6} \times \Bigg(\log_{e}{\bigg|\dfrac{v-3}{v+3}\bigg|}+c_2\Bigg)$

$=\,\,\,$ $\dfrac{1 \times 1}{2 \times 6} \times \Bigg(\log_{e}{\bigg|\dfrac{v-3}{v+3}\bigg|}+c_2\Bigg)$

$=\,\,\,$ $\dfrac{1}{12} \times \Bigg(\log_{e}{\bigg|\dfrac{v-3}{v+3}\bigg|}+c_2\Bigg)$

Now, replace the variable v by its actual value in the function.

$=\,\,\,$ $\dfrac{1}{12}\Bigg(\log_{e}{\Bigg|\dfrac{x^2-3}{x^2+3}\Bigg|}+c_2\Bigg)$

$\therefore\,\,\,$ $I_2$ $\,=\,$ $\dfrac{1}{12}\Bigg(\log_{e}{\Bigg|\dfrac{x^2-3}{x^2+3}\Bigg|}+c_2\Bigg)$

Evaluate the Integration of the given function

In the first step, the integral of function is split as the sum of the integrals of two functions.

$\displaystyle \int{\dfrac{x^{\displaystyle 3}+x}{x^{\displaystyle 4}-9}}\,dx$ $\,=\,$ $\displaystyle \int{\dfrac{x^{\displaystyle 3}}{x^{\displaystyle 4}-9}}\,dx$ $+$ $\displaystyle \int{\dfrac{x}{x^{\displaystyle 4}-9}}\,dx$

$\implies$ $\displaystyle \int{\dfrac{x^{\displaystyle 3}+x}{x^{\displaystyle 4}-9}}\,dx$ $\,=\,$ $I_{1}$ $+$ $I_{2}$

Now, substitute them in right hand side expression to find the integral of the given function.

$\implies$ $\displaystyle \int{\dfrac{x^{\displaystyle 3}+x}{x^{\displaystyle 4}-9}}\,dx$ $\,=\,$ $\dfrac{1}{4}\Big(\log_{e}{\big|x^4-9\big|}+c_1\Big)$ $+$ $\dfrac{1}{12}\Bigg(\log_{e}{\Bigg|\dfrac{x^2-3}{x^2+3}\Bigg|}+c_2\Bigg)$

The factor in each term can be distributed over addition of the terms by the distributive property of multiplication across the addition.

$=\,\,\,$ $\dfrac{1}{4} \times \log_{e}{\big|x^4-9\big|}$ $+$ $\dfrac{1}{4} \times c_1$ $+$ $\dfrac{1}{12} \times \log_{e}{\Bigg|\dfrac{x^2-3}{x^2+3}\Bigg|}$ $+$ $\dfrac{1}{12}\times c_2$

$=\,\,\,$ $\dfrac{1}{4} \times \log_{e}{\big|x^4-9\big|}$ $+$ $\dfrac{1}{12} \times \log_{e}{\Bigg|\dfrac{x^2-3}{x^2+3}\Bigg|}$ $+$ $\dfrac{1}{4}\,c_1$ $+$ $\dfrac{1}{12}\,c_2$

The sum of the third and fourth terms is a constant. Hence, it is simply represented by a constant $c$.

$=\,\,\,$ $\dfrac{1}{4} \times \log_{e}{\big|x^4-9\big|}$ $+$ $\dfrac{1}{12} \times \log_{e}{\Bigg|\dfrac{x^2-3}{x^2+3}\Bigg|}$ $+$ $c$

This expression can be simplified further by expressing the difference of squares in first term in factor form.

$=\,\,\,$ $\dfrac{1}{4} \times \log_{e}{\big|\big(x^2\big)^2-3^2\big|}$ $+$ $\dfrac{1}{12} \times \log_{e}{\Bigg|\dfrac{x^2-3}{x^2+3}\Bigg|}$ $+$ $c$

$=\,\,\,$ $\dfrac{1}{4} \times \log_{e}{\big|(x^2+3)(x^2-3)\big|}$ $+$ $\dfrac{1}{12} \times \log_{e}{\Bigg|\dfrac{x^2-3}{x^2+3}\Bigg|}$ $+$ $c$

Use product rule and quotient rule to split both terms in the expression respectively.

$=\,\,\,$ $\dfrac{1}{4} \times \Big(\log_{e}{\big|x^2+3\big|}+\log_{e}{\big|x^2-3\big|}\Big)$ $+$ $\dfrac{1}{12} \times \Big(\log_{e}{\big|x^2-3\big|}-\log_{e}{\big|x^2+3\big|}\Big)$ $+$ $c$

Now, distribute the multiplying factor over addition and subtraction of the terms to simplify the expression further.

$=\,\,\,$ $\dfrac{1}{4} \times \log_{e}{\big|x^2+3\big|}$ $+$ $\dfrac{1}{4} \times \log_{e}{\big|x^2-3\big|}$ $+$ $\dfrac{1}{12} \times \log_{e}{\big|x^2-3\big|}$ $-$ $\dfrac{1}{12} \times \log_{e}{\big|x^2+3\big|}$ $+$ $c$

$=\,\,\,$ $\dfrac{1}{4} \times \log_{e}{\big|x^2+3\big|}$ $-$ $\dfrac{1}{12} \times \log_{e}{\big|x^2+3\big|}$ $+$ $\dfrac{1}{4} \times \log_{e}{\big|x^2-3\big|}$ $+$ $\dfrac{1}{12} \times \log_{e}{\big|x^2-3\big|}$ $+$ $c$

$=\,\,\,$ $\log_{e}{\big|x^2+3\big|} \times \bigg(\dfrac{1}{4}-\dfrac{1}{12}\bigg)$ $+$ $\log_{e}{\big|x^2-3\big|} \times \bigg(\dfrac{1}{4}+\dfrac{1}{12}\bigg)$ $+$ $c$

$=\,\,\,$ $\bigg(\dfrac{1}{4}-\dfrac{1}{12}\bigg) \times \log_{e}{\big|x^2+3\big|}$ $+$ $\bigg(\dfrac{1}{4}+\dfrac{1}{12}\bigg) \times \log_{e}{\big|x^2-3\big|}$ $+$ $c$

Finally, use the addition and subtraction of the fractions to find the integral of the given algebraic function in rational form.

$=\,\,\,$ $\dfrac{1 \times 3-1 \times 1}{12} \times \log_{e}{\big|x^2+3\big|}$ $+$ $\dfrac{1 \times 3+1 \times 1}{12} \times \log_{e}{\big|x^2-3\big|}$ $+$ $c$

$=\,\,\,$ $\dfrac{3-1}{12} \times \log_{e}{\big|x^2+3\big|}$ $+$ $\dfrac{3+1}{12} \times \log_{e}{\big|x^2-3\big|}$ $+$ $c$

$=\,\,\,$ $\dfrac{2}{12} \times \log_{e}{\big|x^2+3\big|}$ $+$ $\dfrac{4}{12} \times \log_{e}{\big|x^2-3\big|}$ $+$ $c$

$=\,\,\,$ $\dfrac{\cancel{2}}{\cancel{12}} \times \log_{e}{\big|x^2+3\big|}$ $+$ $\dfrac{\cancel{4}}{\cancel{12}} \times \log_{e}{\big|x^2-3\big|}$ $+$ $c$

$=\,\,\,$ $\dfrac{1}{6} \times \log_{e}{\big|x^2+3\big|}$ $+$ $\dfrac{1}{3} \times \log_{e}{\big|x^2-3\big|}$ $+$ $c$

$=\,\,\,$ $\dfrac{1}{6}\,\log_{e}{\big|x^2+3\big|}$ $+$ $\dfrac{1}{3}\,\log_{e}{\big|x^2-3\big|}$ $+$ $c$