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# Evaluate the following integral $\int{\dfrac{{{x}^{5}}}{{{x}^{2}}+9}dx}$

Last updated date: 15th Jul 2024
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Hint: To solve the given integral $\int{\dfrac{{{x}^{5}}}{{{x}^{2}}+9}dx}$, we first have to divide the numerator ${{x}^{5}}$ by the denominator ${{x}^{2}}+9$ using the long division method. Then noting the values of the quotient $q\left( x \right)$ and the remainder $r\left( x \right)$, we will write the numerator as ${{x}^{5}}=\left( {{x}^{2}}+9 \right)q\left( x \right)+r\left( x \right)$ and our integral will be split into two integrals. Then we have to solve the two integrals separately using the basic rules of integration.

Let us write the integral given in the question as
$\Rightarrow I=\int{\dfrac{{{x}^{5}}}{{{x}^{2}}+9}dx}........(i)$
Now we divide the numerator ${{x}^{5}}$ by the denominator ${{x}^{2}}+9$ as below.
{{x}^{2}}+9\overset{{{x}^{3}}-9x}{\overline{\left){\begin{align} & {{x}^{5}} \\ & \underline{{{x}^{5}}+9{{x}^{3}}} \\ & -9{{x}^{3}} \\ & \underline{-9{{x}^{3}}-81x} \\ & \underline{81x} \\ \end{align}}\right.}}
From the above division, the quotient is
$\Rightarrow q\left( x \right)={{x}^{3}}-9x........(ii)$
And the remainder is
$\Rightarrow r\left( x \right)=81x........(iii)$
So the numerator can be written as
$\Rightarrow {{x}^{5}}=\left( {{x}^{2}}+9 \right)q\left( x \right)+r\left( x \right)$
Putting (ii) and (iii) in the above equation, we get
$\Rightarrow {{x}^{5}}=\left( {{x}^{2}}+9 \right)\left( {{x}^{3}}-9x \right)+81x$
Putting the above equation in the equation (i) we get
\begin{align} & \Rightarrow I=\int{\dfrac{\left( {{x}^{2}}+9 \right)\left( {{x}^{3}}-9x \right)+81x}{{{x}^{2}}+9}dx} \\ & \Rightarrow I=\int{\left[ \left( {{x}^{3}}-9x \right)+\dfrac{81x}{{{x}^{2}}+9} \right]dx} \\ & \Rightarrow I=\int{\left( {{x}^{3}}-9x \right)dx}+\int{\dfrac{81x}{{{x}^{2}}+9}dx} \\ \end{align}
Let ${{I}_{1}}=\int{\left( {{x}^{3}}-9x \right)dx}$ and ${{I}_{2}}=\int{\dfrac{81x}{{{x}^{2}}+9}dx}$. So the above equation can be written as
$\Rightarrow I={{I}_{1}}+{{I}_{2}}........(iv)$
Let us solve the first integral.
\begin{align} & \Rightarrow {{I}_{1}}=\int{\left( {{x}^{3}}-9x \right)dx} \\ & \Rightarrow {{I}_{1}}=\int{{{x}^{3}}dx}-\int{9xdx} \\ & \Rightarrow {{I}_{1}}=\int{{{x}^{3}}dx}-9\int{xdx} \\ & \Rightarrow {{I}_{1}}=\dfrac{{{x}^{4}}}{4}-\dfrac{9{{x}^{2}}}{2}+{{C}_{1}}........(v) \\ \end{align}
Now, we solve the second integral.
\begin{align} & \Rightarrow {{I}_{2}}=\int{\dfrac{81x}{{{x}^{2}}+9}dx} \\ & \Rightarrow {{I}_{2}}=81\int{\dfrac{xdx}{{{x}^{2}}+9}}........(vi) \\ \end{align}
Let us substitute $t={{x}^{2}}+9$ so that
$\Rightarrow {{x}^{2}}+9=t$
Differentiating both the sides with respect to $x$ we get
$\Rightarrow 2x=\dfrac{dt}{dx}$
Multiplying by $dx$ both sides
\begin{align} & \Rightarrow 2xdx=dt \\ & \Rightarrow xdx=\dfrac{dt}{2} \\ \end{align}
Substituting this in (vi) we get
\begin{align} & \Rightarrow {{I}_{2}}=81\int{\dfrac{dt}{2t}} \\ & \Rightarrow {{I}_{2}}=\dfrac{81}{2}\int{\dfrac{dt}{t}} \\ & \Rightarrow {{I}_{2}}=\dfrac{81}{2}\ln \left| t \right|+{{C}_{2}} \\ \end{align}
Back substituting $t={{x}^{2}}+9$ we get
$\Rightarrow {{I}_{2}}=\dfrac{81}{2}\ln \left| {{x}^{2}}+9 \right|+{{C}_{2}}$
Since ${{x}^{2}}+9>0$ for all $x$, we can write $\left| {{x}^{2}}+9 \right|={{x}^{2}}+9$ in the above equation to get
$\Rightarrow {{I}_{2}}=\dfrac{81}{2}\ln \left( {{x}^{2}}+9 \right)+{{C}_{2}}........(vii)$
Substituting (v) and (vii) in (iv) we get
\begin{align} & \Rightarrow I=\dfrac{{{x}^{4}}}{4}-\dfrac{9{{x}^{2}}}{2}+{{C}_{1}}+\dfrac{81}{2}\ln \left( {{x}^{2}}+9 \right)+{{C}_{2}} \\ & \Rightarrow I=\dfrac{{{x}^{4}}}{4}-\dfrac{9{{x}^{2}}}{2}+\dfrac{81}{2}\ln \left( {{x}^{2}}+9 \right)+{{C}_{1}}+{{C}_{2}} \\ & \Rightarrow I=\dfrac{{{x}^{4}}}{4}-\dfrac{9{{x}^{2}}}{2}+\dfrac{81}{2}\ln \left( {{x}^{2}}+9 \right)+C \\ \end{align}
Where $C={{C}_{1}}+{{C}_{2}}$

Hence, the given integral is equal to $\dfrac{{{x}^{4}}}{4}-\dfrac{9{{x}^{2}}}{2}+\dfrac{81}{2}\ln \left( {{x}^{2}}+9 \right)+C$.

Note: We can also solve this question without performing the division by directly substituting the denominator ${{x}^{2}}+9=t$ and writing the given integral as $\int{\dfrac{{{x}^{4}}xdx}{{{x}^{2}}+9}}$ and putting ${{x}^{4}}={{\left( t-1 \right)}^{2}}$ and $xdx=\dfrac{dt}{2}$. In this case, our integral will become $\dfrac{1}{2}\int{\dfrac{{{\left( t-9 \right)}^{2}}}{t}}dt$ which can be easily solved by expanding the numerator. By this method too, we will get the same answer as that in the above solution.