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

& I={{I}_{1}}+{{I}_{2}} \\\ & \Rightarrow I=0+{{I}_{2}} \\\ \end{aligned}$$ $$\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{aligned} & =\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{aligned}$$ $$={{x}^{4}}-2{{x}^{2}}+\dfrac{4{{x}^{2}}}{\left( {{x}^{2}}+2 \right)}$$ …………………….(9) From equation (8) and equation (9), we get $$\begin{aligned} & {{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{aligned}$$ $$\begin{aligned} & {{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{aligned}$$ $$\begin{aligned} & {{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{aligned}$$ From equation (6), we have $${{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}$$