Question

Evaluate the definite integral, \[\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{2{{x}^{7}}+3{{x}^{6}}-10{{x}^{5}}-7{{x}^{3}}-12{{x}^{2}}+x+1}{{{x}^{2}}+2}dx}\] .

Answer 1

Hint: Assume \[I=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{2{{x}^{7}}+3{{x}^{6}}-10{{x}^{5}}-7{{x}^{3}}-12{{x}^{2}}+x+1}{{{x}^{2}}+2}dx}\] where \[{{I}_{1}}=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{2{{x}^{7}}-10{{x}^{5}}-7{{x}^{3}}+x}{{{x}^{2}}+2}dx}\] and \[{{I}_{2}}=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{3{{x}^{6}}-12{{x}^{2}}+1}{{{x}^{2}}+2}dx}\] . We also know that \[f(x)={{x}^{n}}\] is an odd function if n is odd. We know the property, \[\int_{-a}^{a}{f(x)dx}=0\] where \[f(x)\] is an odd function. So, the value of \[{{I}_{1}}\] is 0. We also know that \[f(x)={{x}^{n}}\] is an odd function if n is even. We know the property, \[\int_{-a}^{a}{f(x)dx}=2\int_{0}^{a}{f(x)dx}\] where \[f(x)\] is an even function. Using this property simplifies \[{{I}_{2}}\] . Transform \[{{I}_{2}}\] as \[{{I}_{2}}=3\int\limits_{0}^{\sqrt{2}}{\dfrac{{{x}^{6}}}{{{x}^{2}}+2}dx-24\int\limits_{0}^{\sqrt{2}}{\dfrac{{{x}^{2}}}{{{x}^{2}}+2}dx+2\int\limits_{0}^{\sqrt{2}}{\dfrac{1}{{{x}^{2}}+{{\left( \sqrt{2} \right)}^{2}}}dx}}}\] . In this equation of \[{{I}_{2}}\] , we can write the term \[\left( \dfrac{{{x}^{6}}}{{{x}^{2}}+2} \right)\] as \[\left( {{x}^{4}}-2{{x}^{2}}+\dfrac{4{{x}^{2}}}{{{x}^{2}}+2} \right)\] . Use this, simplify and solve it further.

Complete step-by-step answer:
Let \[I=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{2{{x}^{7}}+3{{x}^{6}}-10{{x}^{5}}-7{{x}^{3}}-12{{x}^{2}}+x+1}{{{x}^{2}}+2}dx}\] where \[{{I}_{1}}=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{2{{x}^{7}}-10{{x}^{5}}-7{{x}^{3}}+x}{{{x}^{2}}+2}dx}\] and
\[{{I}_{2}}=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{3{{x}^{6}}-12{{x}^{2}}+1}{{{x}^{2}}+2}dx}\] .
\[I=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{2{{x}^{7}}+3{{x}^{6}}-10{{x}^{5}}-7{{x}^{3}}-12{{x}^{2}}+x+1}{{{x}^{2}}+2}dx}\] ………………………….(1)
\[{{I}_{1}}=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{2{{x}^{7}}-10{{x}^{5}}-7{{x}^{3}}+x}{{{x}^{2}}+2}dx}\] ……………………………..(2)
\[{{I}_{2}}=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{3{{x}^{6}}-12{{x}^{2}}+1}{{{x}^{2}}+2}dx}\] ……………………………(3)
We can now say that \[I\] is the summation of \[{{I}_{1}}\] and \[{{I}_{2}}\] .
\[I={{I}_{1}}+{{I}_{2}}\] ……………………..(4)
We also know that \[f(x)={{x}^{n}}\] is an odd function if n is odd.
In equation (2) we have \[{{x}^{7}},{{x}^{5}},{{x}^{3}},x\] and these all have odd numbers as their exponents. So, these all are odd functions.
We know the property, \[\int_{-a}^{a}{f(x)dx}=0\] where \[f(x)\] is an odd function.
Now, using the property we can say that, \[{{I}_{1}}=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{2{{x}^{7}}-10{{x}^{5}}-7{{x}^{3}}+x}{{{x}^{2}}+2}dx}\] is equal to zero. So,
\[{{I}_{1}}=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{2{{x}^{7}}-10{{x}^{5}}-7{{x}^{3}}+x}{{{x}^{2}}+2}dx}=0\] ………………………..(5)
Now, putting the value of \[{{I}_{1}}\] in equation (4), we get
\[\begin{align}
  & I={{I}_{1}}+{{I}_{2}} \\
 & \Rightarrow I=0+{{I}_{2}} \\
\end{align}\]
\[\Rightarrow I={{I}_{2}}\] ………………(6)
We also know that \[f(x)={{x}^{n}}\] is an odd function if n is even.
In equation (3) we have \[{{x}^{6}}\,\text{and }{{x}^{2}}\] , and these all have even numbers as their exponents. So, these all are even functions.
We know the property, \[\int_{-a}^{a}{f(x)dx}=2\int_{0}^{a}{f(x)dx}\] where \[f(x)\] is an even function.
Using this property for simplifying equation (3), we get
\[{{I}_{2}}=2\int\limits_{0}^{\sqrt{2}}{\dfrac{3{{x}^{6}}-12{{x}^{2}}+1}{{{x}^{2}}+2}dx}\]
\[{{I}_{2}}=2\int\limits_{0}^{\sqrt{2}}{\dfrac{3{{x}^{6}}}{{{x}^{2}}+2}dx-24\int\limits_{0}^{\sqrt{2}}{\dfrac{{{x}^{2}}}{{{x}^{2}}+2}dx+2\int\limits_{0}^{\sqrt{2}}{\dfrac{1}{{{x}^{2}}+{{\left( \sqrt{2} \right)}^{2}}}dx}}}\] ……………………..(7)
We know the formula, \[\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)}\] .
Using this formula and transforming equation (7), we get
\[{{I}_{2}}=2{{\int\limits_{0}^{\sqrt{2}}{\dfrac{3{{x}^{6}}}{{{x}^{2}}+2}dx-24\int\limits_{0}^{\sqrt{2}}{\dfrac{{{x}^{2}}}{{{x}^{2}}+2}dx+2.\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{2}} \right)}}}_{0}}^{\sqrt{2}}\]
We know that \[{{\tan }^{-1}}1=\dfrac{\pi }{4}\,\text{and ta}{{\text{n}}^{-1}}0=0\] .
\[{{I}_{2}}=2.3\int_{0}^{\sqrt{2}}{\dfrac{{{x}^{6}}}{{{x}^{2}}+2}dx}-24\int_{0}^{\sqrt{2}}{\dfrac{{{x}^{2}}}{{{x}^{2}}+2}dx+}\sqrt{2}\left( {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right)\]
\[{{I}_{2}}=6\int_{0}^{\sqrt{2}}{\dfrac{{{x}^{6}}}{{{x}^{2}}+2}dx}-24\int_{0}^{\sqrt{2}}{\dfrac{{{x}^{2}}}{{{x}^{2}}+2}dx+}\sqrt{2}\left( \dfrac{\pi }{4}-0 \right)\] …………………………….(8)
Now, simplifying the term \[\left( \dfrac{{{x}^{6}}}{{{x}^{2}}+2} \right)\] in equation (8), we get
\[\begin{align}
  & =\dfrac{{{x}^{6}}}{{{x}^{2}}+2} \\
 & =\dfrac{{{x}^{6}}+2{{x}^{4}}-2{{x}^{4}}-4{{x}^{2}}+4{{x}^{2}}}{{{x}^{2}}+2} \\
 & =\dfrac{{{x}^{4}}\left( {{x}^{2}}+2 \right)-2{{x}^{2}}\left( {{x}^{2}}+2 \right)+4{{x}^{2}}}{{{x}^{2}}+2} \\
 & =\dfrac{{{x}^{4}}\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+2 \right)}-\dfrac{2{{x}^{2}}\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+2 \right)}+\dfrac{4{{x}^{2}}}{\left( {{x}^{2}}+2 \right)} \\
\end{align}\]
\[={{x}^{4}}-2{{x}^{2}}+\dfrac{4{{x}^{2}}}{\left( {{x}^{2}}+2 \right)}\] …………………….(9)
From equation (8) and equation (9), we get
\[\begin{align}
  & {{I}_{2}}=6\int_{0}^{\sqrt{2}}{\left( {{x}^{4}}-2{{x}^{2}}+\dfrac{4{{x}^{2}}}{{{x}^{2}}+2} \right)}dx-24\int_{0}^{\sqrt{2}}{\dfrac{{{x}^{2}}}{{{x}^{2}}+2}dx+}\sqrt{2}.\dfrac{\pi }{4} \\
 & {{I}_{2}}=6\int_{0}^{\sqrt{2}}{\left( {{x}^{4}}-2{{x}^{2}} \right)}dx+6\int_{0}^{\sqrt{2}}{\dfrac{4{{x}^{2}}}{{{x}^{2}}+2}d}x-24\int_{0}^{\sqrt{2}}{\dfrac{{{x}^{2}}}{{{x}^{2}}+2}dx+}\dfrac{\pi }{2\sqrt{2}} \\
 & {{I}_{2}}=6\int_{0}^{\sqrt{2}}{\left( {{x}^{4}}-2{{x}^{2}} \right)}dx+24\int_{0}^{\sqrt{2}}{\dfrac{{{x}^{2}}}{{{x}^{2}}+2}d}x-24\int_{0}^{\sqrt{2}}{\dfrac{{{x}^{2}}}{{{x}^{2}}+2}dx+}\dfrac{\pi }{2\sqrt{2}} \\
\end{align}\]
\[\begin{align}
  & {{I}_{2}}=6{{\left[ \dfrac{{{x}^{5}}}{5}-\dfrac{2{{x}^{3}}}{3} \right]}_{0}}^{\sqrt{2}}+\dfrac{\pi }{2\sqrt{2}} \\
 & {{I}_{2}}=6\left[ \dfrac{4\sqrt{2}}{5}-\dfrac{4\sqrt{2}}{3} \right]+\dfrac{\pi }{2\sqrt{2}} \\
\end{align}\]
\[\begin{align}
  & {{I}_{2}}=6\times 4\sqrt{2}\left( \dfrac{1}{5}-\dfrac{1}{3} \right)+\dfrac{\pi }{2\sqrt{2}} \\
 & {{I}_{2}}=24\sqrt{2}\left( \dfrac{3-5}{15} \right)+\dfrac{\pi }{2\sqrt{2}} \\
 & {{I}_{2}}=24\sqrt{2}\left( \dfrac{-2}{15} \right)+\dfrac{\pi }{2\sqrt{2}} \\
 & {{I}_{2}}=\dfrac{-48\sqrt{2}}{15}+\dfrac{\pi }{2\sqrt{2}} \\
 & {{I}_{2}}=\dfrac{-16\sqrt{2}}{5}+\dfrac{\pi }{2\sqrt{2}} \\
\end{align}\]
From equation (6), we have \[{{I}_{2}}\] is equal to \[I\] .
So, \[I={{I}_{2}}=\dfrac{-16\sqrt{2}}{5}+\dfrac{\pi }{2\sqrt{2}}\] .
From equation (1), we have \[I=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{2{{x}^{7}}+3{{x}^{6}}-10{{x}^{5}}-7{{x}^{3}}-12{{x}^{2}}+x+1}{{{x}^{2}}+2}dx}\] .
Hence, \[I=\int\limits_{-\sqrt{2}}^{\sqrt{2}}{\dfrac{2{{x}^{7}}+3{{x}^{6}}-10{{x}^{5}}-7{{x}^{3}}-12{{x}^{2}}+x+1}{{{x}^{2}}+2}dx}=\dfrac{-16\sqrt{2}}{5}+\dfrac{\pi }{2\sqrt{2}}\] .

Note: To solve this question, one may think to factorise and try to get the term \[({{x}^{2}}+2)\] as a factor in the numerator of \[\dfrac{2{{x}^{7}}+3{{x}^{6}}-10{{x}^{5}}-7{{x}^{3}}-12{{x}^{2}}+x+1}{{{x}^{2}}+2}\] so that the term \[({{x}^{2}}+2)\] gets cancelled from the denominator. But this approach cannot work here. We cannot make the term \[({{x}^{2}}+2)\] as a factor in the numerator of \[\dfrac{2{{x}^{7}}+3{{x}^{6}}-10{{x}^{5}}-7{{x}^{3}}-12{{x}^{2}}+x+1}{{{x}^{2}}+2}\]