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.

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.

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

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 $u$.

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

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

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.

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

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

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.

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

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

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

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

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

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

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