Mathematics

# $I = \int \dfrac {1}{x(x^{6}+1)}dx$

##### SOLUTION
$I = \int {\dfrac{{\rm{1}}}{{x\left( {{x^{\rm{6}}} + {\rm{1}}} \right)}}{\rm{dx}}}$
putting $x^6+1=t$
$6x^5dx=dt$
$dx=\dfrac{dt}{6x^5}$
$I = \int {\dfrac{{dt}}{{6{x^5} \times x \times t}}}$
$I = \dfrac{1}{6}\int {\frac{{dt}}{{t\left( {t - 1} \right)}}}$
$I = \dfrac{1}{6}\int {\left( {\dfrac{1}{{t - 1}} - \dfrac{1}{t}} \right)dt}$
$I = \dfrac{1}{6}\int {\frac{1}{{t - 1}}dt} - \dfrac{1}{6}\int {\dfrac{1}{t}dt}$
$I =\dfrac{1}{6}\log (t-1)-\dfrac{1}{6}\log (t)+C$
$I =\dfrac{1}{6}\log (\dfrac{t-1}{t})+C$
$I =\dfrac{1}{6}\log (\dfrac{x^6}{x^6+1})+C$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Subjective Medium
Evaluate $\displaystyle\int_{0}^{2\pi}\cos^{-1}(\cos x)\ dx$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Evaluate $\displaystyle\int{\frac{dx}{\sqrt{(x-a)(b-x)}}}$.
• A. $\displaystyle I=2\cos^{-1}{\sqrt{\frac{x-a}{(b-a)}}}+C$
• B. $\displaystyle I=\sin^{-1}{\sqrt{\frac{x-a}{(b-a)}}}+C$
• C. $\displaystyle I=2\sin^{-1}{\sqrt{\frac{x-b}{(a-b)}}}+C$
• D. $\displaystyle I=2\sin^{-1}{\sqrt{\frac{x-a}{(b-a)}}}+C$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard

If the integral $\displaystyle \int\frac{5\tan x}{\tan x-2}dx=x+$ $a\ ln$ $|\sin x-2\cos \mathrm{x}|+k$ then $a$ is equal to:
• A. $1$
• B. $-2$
• C. $-1$
• D. $2$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Multiple Correct Hard

What is the geometrical interpretation of indefinite integral?

• A. These members cannot be obtained by shifting any one of the curves parallel to itself.
• B. None of these
• C. The equation $\displaystyle\int f(x){dx}= F(x)+C$, represents a family of curves.
• D. The different values of $C$ will correspond to different members of this family.

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Medium
The average value of a function f(x) over the interval, [a,b] is the number $\displaystyle \mu =\frac{1}{b-a}\int_{a}^{b}f\left ( x \right )dx$
The square root $\displaystyle \left \{ \frac{1}{b-a}\int_{a}^{b}\left [ f\left ( x \right ) \right ]^{2}dx \right \}^{1/2}$ is called the root mean square of f on [a, b]. The average value of $\displaystyle \mu$ is attained id f is continuous on [a, b].

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020