Mathematics

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

##### SOLUTION

Consider the given integral.

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

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

We know that

$\int{\dfrac{dx}{1+{{x}^{2}}}={{\tan }^{-1}}x+C}$

Therefore,

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

$I=\dfrac{{{x}^{5}}}{5}-\dfrac{{{x}^{3}}}{3}+x-2{{\tan }^{-1}}x+C$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 Single Correct Medium
If $\displaystyle f\left ( x \right )= \left\{\begin{matrix}e^{\cos x}\cdot \sin x &for\left | x \right |\leq 2 \\2 &otherwise \end{matrix}\right.$ then $\displaystyle \int_{-2}^{3}f\left ( x \right )dx$ is equal to
• A.
• B. 1
• C. 3
• D. 2

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the given integral: $\displaystyle \int_{0}^{5}x^4\ dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 One Word Medium
Solve $\displaystyle \int_{0}^{\pi /2}\frac{\cos x-\sin x}{1+\sin x\cos x}dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve:
$\int_{}^{} {\frac{{\cos x}}{{\sqrt {{{\sin }^2}x - 2\sin x - 3} }}} dx$

$\displaystyle \int^{\dfrac{\pi}{2}}_0\dfrac{\sin x}{\sin x+\cos x}dx$