Mathematics

# $\int{ \dfrac {1} {{ x }^{ 2 } - {x} + {1}} dx}$

$= \dfrac {2}{\sqrt3}\tan^{-1} (\dfrac{2x-1}{\sqrt3})+C$

##### SOLUTION
$\int{ \dfrac {1} {{ x }^{ 2 } - {x} + {1}} dx}$
$= \int{ \dfrac {1} {{ (x- \dfrac{1}{2}})^{ 2 } + \dfrac {3}{4}} dx}$
$= \int{ \dfrac {1} {{ (x- \dfrac{1}{2}})^{ 2 } + (\dfrac {\sqrt3}{2})^2} dx}$
$= \dfrac {1}{\dfrac {\sqrt{3}}{2}}{\tan^{-1}}(\dfrac{x-\dfrac 12}{\dfrac{\sqrt3}2})+C$
$= \dfrac {2}{\sqrt3}\tan^{-1} (\dfrac{2x-1}{\sqrt3})+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
$\int \sqrt { 1 - \sin x } d x =$
• A. $2\sqrt {1-\sin x}+C$
• B. $2\sqrt {1-2\sin x}+C$
• C. $2\sqrt {1-\sin 2x}+C$
• D. $2\sqrt {1+\sin x}+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
If $\displaystyle \int \dfrac {4e^{x}+6e^{-x}}{9e^{x}+4e^{-x}}dx=Ax+B\log (9e^{x}+4^{-x})+C$, then
• A. $A=\dfrac {19}{36},B=\dfrac {35}{36},C=0$
• B. $A=\dfrac {-19}{36},B=\dfrac {35}{36},C=0$
• C. $A=\dfrac {-19}{36},B=\dfrac {-35}{36},C\in \ R$
• D. $A=\dfrac {35}{36},B=\dfrac {-19}{36},C\in \ R$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Show that $\displaystyle \int \sqrt{\left ( \frac{x+1}{x-1} \right )}dx=\sqrt{x^{2}-1}+\log[x+\sqrt{x^{2}-1}]$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If  $\displaystyle I_n=\int_{0}^{\tfrac{\pi}{4}} \tan^nx\sec^2xdx,$ then  $I_1, I_2, I_3..$ are  in
• A. A.P.
• B. G.P.
• C. A.G.P.
• D. H.P.

Let $n \space\epsilon \space N$ & the A.M., G.M., H.M. & the root mean square of $n$ numbers $2n+1, 2n+2, ...,$ up to $n^{th}$ number are $A_{n}$, $G_{n}$, $H_{n}$ and $R_{n}$ respectively.