How do you find the integral of $\frac{6 x + 1}{x^{2} + 2 x + 3}$ $(6x+1)/(x^2+2x+3)$ ?

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How do you find the integral of $\frac{6 x + 1}{x^{2} + 2 x + 3}$ $(6x+1)/(x^2+2x+3)$ ?

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Answer 1 · 2018-03-18T17:53:23

$I = 3 ln ∣ ∣ x^{2} + 2 x + 3 ∣ ∣ - \frac{5}{\sqrt{2}} {tan}^{- 1} (\frac{x + 1}{\sqrt{2}}) + c$ $I=3ln|x^2+2x+3|-5/sqrt(2)tan^-1((x+1)/sqrt(2))+c$

Explanation:

$I = \int \frac{6 x + 1}{x^{2} + 2 x + 3} d x$ $I=int(6x+1)/(x^2+2x+3)dx$

$= \int \frac{6 x + 6}{x^{2} + 2 x + 3} d x - \int \frac{5}{x^{2} + 2 x + 3} d x$ $=int(6x+6)/(x^2+2x+3)dx-int5/(x^2+2x+3)dx$

$I = 3 \int \frac{2 x + 2}{x^{2} + 2 x + 3} d x - \int \frac{5}{x^{2} + 2 x + 1 + 2} d x$ $I=3int(2x+2)/(x^2+2x+3)dx-int5/(x^2+2x+1+2)dx$

$I = 3 \int \frac{\frac{d}{d x} (x^{2} + 2 x + 3)}{x^{2} + 2 x + 3} d x - \int \frac{5}{{(x + 1)}^{2} + {(\sqrt{2})}^{2}} d x$ $I=3int(d/(dx)(x^2+2x+3))/(x^2+2x+3)dx-int5/((x+1)^2+(sqrt(2))^2)dx$

$I = 3 ln ∣ ∣ x^{2} + 2 x + 3 ∣ ∣ - 5 \cdot \frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{x + 1}{\sqrt{2}}) + c$ $I=3ln|x^2+2x+3|-5*1/sqrt(2)tan^-1((x+1)/sqrt(2))+c$
$I = 3 ln ∣ ∣ x^{2} + 2 x + 3 ∣ ∣ - \frac{5}{\sqrt{2}} {tan}^{- 1} (\frac{x + 1}{\sqrt{2}}) + c$ $I=3ln|x^2+2x+3|-5/sqrt(2)tan^-1((x+1)/sqrt(2))+c$

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How do you find the integral of $\frac{6 x + 1}{x^{2} + 2 x + 3}$ $(6x+1)/(x^2+2x+3)$ ?

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How do you find the integral of (6x+1)/(x^2+2x+3)6x+1x2+2x+3(6x+1)/(x^2+2x+3)?

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Explanation:

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How do you find the integral of $\frac{6 x + 1}{x^{2} + 2 x + 3}$ $(6x+1)/(x^2+2x+3)$ ?