Mathematics

# Solve $\displaystyle\int {{{\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)}^6}dx}$

##### SOLUTION

Consider the given integral.

$I=\displaystyle\int{{{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{6}}}dx$

$I=\displaystyle\int{{{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}{{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}{{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{2}}}dx$

$I=\displaystyle\int{\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}}+2 \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}}+2 \right)\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}}+2 \right)}dx$

$I=\displaystyle\int{\left( {{x}^{4}}+\dfrac{1}{{{x}^{4}}}+2 \right)\left( {{x}^{8}}+\dfrac{1}{{{x}^{8}}}+4{{x}^{4}}+\dfrac{4}{{{x}^{4}}}+8 \right)}dx$

$I=\displaystyle\int{\left( {{x}^{12}}+\dfrac{15}{{{x}^{4}}}+\dfrac{6}{{{x}^{8}}}+\dfrac{1}{{{x}^{12}}}+6{{x}^{8}}+15{{x}^{4}}+20 \right)}dx$

$I=\left( \dfrac{{{x}^{13}}}{13}-\dfrac{6}{{{x}^{3}}}-\dfrac{6}{7{{x}^{7}}}-\dfrac{1}{11{{x}^{11}}}+\dfrac{2{{x}^{9}}}{3}+3{{x}^{5}}+20x \right)+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
$\int_0^{\pi/2} \dfrac{tan^7 x}{cot^7 x + tan^7 x} dx$ is equal to
• A. $\dfrac{\pi}{6}$
• B. $\dfrac{\pi}{2}$
• C. $\dfrac{\pi}{3}$
• D. $\dfrac{\pi}{4}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int_{-1}^{1 / 2} \dfrac{e^{x}\left(2-x^{2}\right) d x}{(1-x) \sqrt{1-x^{2}}}$ is equal to
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• C. $\sqrt{\dfrac{e}{3}}$
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1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
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Q4 Subjective Medium
Evaluate the following integral
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