Mathematics

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

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

$\displaystyle = \frac{1}{2}\int \frac{2x}{x^{2}+x+1}$
$\displaystyle = \frac{1}{2}\int \frac{2x+1-1}{x^{2}+x+1}dx$

$\displaystyle = \frac{1}{2}\int \frac{2x+1}{x^{2}+x+1}dx-\frac{1}{2}\frac{1}{x^{2}+x+1}dx$

$\displaystyle = \frac{1}{2}log (x^{2}+x+1)-\frac{1}{2}\int \frac{1}{x^{2}+2.x\frac{1}{2}+(\frac{1}{2})^{2}-(\frac{1}{2})^{2}+1}$

$\displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{2}\int \frac{1}{(x+\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}}dx$

Let $\displaystyle x+\frac{1}{2} = t \Rightarrow dx = dt$

$\displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{2}\int \frac{dt}{t^{2}+(\frac{\sqrt{3}}{2})^{2}}$

$\displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{2}.\frac{1}{(\frac{\sqrt{3}}{2})}ten^{-1}(\frac{t}{(\sqrt{3}/2)})+c$

$\displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{\sqrt{3}}tan^{-1}(\frac{2(x+1/2)}{\sqrt{3}})+c$

$\displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{\sqrt{3}}tan^{-1}(\frac{2x+1}{\sqrt{3}})+c$

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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
$n\overset{Lt}{\rightarrow}\infty [\displaystyle \frac{1}{1-n^{2}}+\frac{2}{1-n^{2}}+\ldots+\frac{n}{1-n^{2}}]=$
• A.
• B. 1/2
• C. 1
• D. -1/2

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value of $\displaystyle \int_{-\frac {\pi}{2}}^{\frac {\pi}{2}}(x^3+x \cos x+\tan^5x+1)dx$
• A. $0$
• B. $2$
• C. $1$
• D. $\pi$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If  $I_{1}=\displaystyle \lim_{n\rightarrow \infty }\sum_{r=1}^{2n} \dfrac{n}{r^{2}+3rn+2n^{2}}$

and $\displaystyle I_{2}=\lim_{n\rightarrow \infty }\sum_{r=1}^{2n} \dfrac{n}{2r^{2}+3rn+n^{2}}$

then the value of $2I_{1} - I_{2}$ is equal to?
• A. $\ln\dfrac{4}{3}$
• B. $\ln\dfrac{3}{4}$
• C. $0$
• D. $\ln\dfrac{3}{2}$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\underset{n - 1}{\overset{100}{\sum}} \underset{n - 1}{\overset{n}{\int}} \, e^{x - [x]} dx$ =
• A. $\dfrac{e - 1}{100}$
• B. $100 (e - 1)$
• C. $\dfrac{e^{100} - 1}{100}$
• D. $\dfrac{e^{100} - 1}{e - 1}$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Subjective Easy
Evaluate:
$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020