Question

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

Answer 1

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.
${{x}^{2}}+9\overset{{{x}^{3}}-9x}{\overline{\left){\begin{aligned} & {{x}^{5}} \\\ & \underline{{{x}^{5}}+9{{x}^{3}}} \\\ & -9{{x}^{3}} \\\ & \underline{-9{{x}^{3}}-81x} \\\ & \underline{81x} \\\ \end{aligned}}\right.}}$
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

& \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{aligned}$$ Let $${{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{aligned} & \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{aligned}$$ Now, we solve the second integral. $\begin{aligned} & \Rightarrow {{I}_{2}}=\int{\dfrac{81x}{{{x}^{2}}+9}dx} \\\ & \Rightarrow {{I}_{2}}=81\int{\dfrac{xdx}{{{x}^{2}}+9}}........(vi) \\\ \end{aligned}$ Let us substitute $t={{x}^{2}}+9$ so that $\Rightarrow {{x}^{2}}+9=t$ Differentiating both the sides with respect to $x$ we get $\Rightarrow 2x=\dfrac{dt}{dx}$ Multiplying by $dx$ both sides $$\begin{aligned} & \Rightarrow 2xdx=dt \\\ & \Rightarrow xdx=\dfrac{dt}{2} \\\ \end{aligned}$$ Substituting this in (vi) we get $$\begin{aligned} & \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{aligned}$$ Back substituting $t={{x}^{2}}+9$ we get $\Rightarrow {{I}_{2}}=\dfrac{81}{2}\ln \left| {{x}^{2}}+9 \right|+{{C}_{2}}$ Since ${{x}^{2}}+9>0$ for all $x$, we can write $\left| {{x}^{2}}+9 \right|={{x}^{2}}+9$ in the above equation to get $$\Rightarrow {{I}_{2}}=\dfrac{81}{2}\ln \left( {{x}^{2}}+9 \right)+{{C}_{2}}........(vii)$$ Substituting (v) and (vii) in (iv) we get $\begin{aligned} & \Rightarrow I=\dfrac{{{x}^{4}}}{4}-\dfrac{9{{x}^{2}}}{2}+{{C}_{1}}+\dfrac{81}{2}\ln \left( {{x}^{2}}+9 \right)+{{C}_{2}} \\\ & \Rightarrow I=\dfrac{{{x}^{4}}}{4}-\dfrac{9{{x}^{2}}}{2}+\dfrac{81}{2}\ln \left( {{x}^{2}}+9 \right)+{{C}_{1}}+{{C}_{2}} \\\ & \Rightarrow I=\dfrac{{{x}^{4}}}{4}-\dfrac{9{{x}^{2}}}{2}+\dfrac{81}{2}\ln \left( {{x}^{2}}+9 \right)+C \\\ \end{aligned}$ 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.