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

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Last updated date: 25th Feb 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.

Complete step by step answer:
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.