Find the ∫ ((x+1)/(x^2 +x+1))dx


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Question Number 32139 by Cheyboy last updated on 20/Mar/18

$F i n d t h e$ $\int \frac{x + 1}{x^{2} + x + 1} d x$
Commented by mrW2 last updated on 20/Mar/18

$\int \frac{x + 1}{x^{2} + x + 1} d x$ $= \frac{1}{2} \int \frac{2 x + 1 + 1}{x^{2} + x + 1} d x$ $= \frac{1}{2} [\int \frac{2 x + 1}{x^{2} + x + 1} d x + \int \frac{1}{x^{2} + x + 1} d x]$ $= \frac{1}{2} [\int \frac{1}{x^{2} + x + 1} d (x^{2} + x + 1) + \int \frac{1}{x^{2} + x + 1} d x]$ $= \frac{1}{2} [\ln (x^{2} + x + 1) + \int \frac{1}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d (x + \frac{1}{2})]$ $= \frac{1}{2} [\ln (x^{2} + x + 1) + \frac{2}{\sqrt{3}} \tan^{- 1} \frac{2 x + 1}{\sqrt{3}}] + C$ $= \frac{1}{2} \ln (x^{2} + x + 1) + \frac{1}{\sqrt{3}} \tan^{- 1} \frac{2 x + 1}{\sqrt{3}} + C$
Commented by Cheyboy last updated on 20/Mar/18

$S i r p l z z c a n u e x p l a i n t h i s p a r t$ $\int \frac{1}{x^{2} + x + 1} d (x^{2} + x + 1)$
Commented by Tinkutara last updated on 20/Mar/18

$\frac{d}{d x} (x^{2} + x + 1) = 2 x + 1$ $d (x^{2} + x + 1) = (2 x + 1) d x$ $(2 x + 1) d x i s r e p l a c e d b y d (x^{2} + x + 1) .$
Commented by Cheyboy last updated on 20/Mar/18

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