Mathematics

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

##### SOLUTION
I=$\int\dfrac{x}{x^2+x+1}dx=\int\dfrac{\dfrac{1}{2}(2x+1)-\dfrac{1}{2}}{x^2+x+1}dx=\dfrac{1}{2}\int\dfrac{2x+1}{x^2+x+1}dx(i_{1})-\dfrac{1}{2}\int\dfrac{1}{x^2+x+1}dx(i_{2})+C$

$i_{1}$ can be read as $\dfrac{1}{2}\int\dfrac{dt}{t}$ for t=$x^2+x+1$
so $i_{1}=\dfrac{1}{2}ln|x^2+x+1|+c$

$i_{2}$ can simply be evaluated  by making perfect square in denominator
$i_{2}=\dfrac{1}{2}\int\dfrac{1}{(x+\dfrac{1}{2})^{2}+\dfrac{3}{4}}=\dfrac{1}{\sqrt{3}}tan^{-1}(\dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}})+c$

So I=$i_{1}-i_{2}=$ $\dfrac{1}{2}ln|x^2+x+1|-\dfrac{1}{\sqrt{3}}tan^{-1}(\dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}})+C$

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

#### Realted Questions

Q1 Subjective Medium
Evaluate the following integral:
$\int { \cfrac { 1 }{ 1+\sqrt { x } } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
If $\displaystyle I=\int \log\left ( \sqrt{1-x}+\sqrt{1+x} \right )dx,$ then I is equal to

• A. $\displaystyle x\log \left ( \sqrt{1-x}+\sqrt{1+x} \right )+\frac{1}{2}x+C$
• B. $\displaystyle x\log \left ( \sqrt{1-x}+\sqrt{1+x} \right )+\frac{1}{2}\sin^{-1}x+C$
• C. $\displaystyle x\log \left ( \sqrt{1-x}+\sqrt{1+x} \right )+\frac{1}{2}\sin^{-1}x+\frac{1}{2}x+C$
• D. $\displaystyle x\log \left ( \sqrt{1-x}+\sqrt{1+x} \right )+\frac{1}{2}\sin^{-1}x-\frac{1}{2}x+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of ${\int}_{0}^{\pi}\left|\sin^{4}x\right|dx$ is
• A. $\dfrac{2\pi}{3}$
• B. $\dfrac{4\pi}{3}$
• C. $\dfrac{3\pi}{3}$
• D. $\dfrac{8\pi}{3}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Show that $\displaystyle \int \frac{x dx}{(px+q)^{3/2}}=\frac{1}{p^{2}}\left \{ \sqrt{px+q}+\frac{q}{\sqrt{(px+q)}} \right \}\cdot$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$