∫((3x+2)/(5x^2 +2x+3))dx=?


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Question Number 213098 by MathematicalUser2357 last updated on 30/Oct/24

$\int \frac{3 x + 2}{5 x^{2} + 2 x + 3} d x = ?$
	Answered by issac last updated on 30/Oct/24

	$Hmmmmm . . . . .$ $\frac{3 x + 2}{5 x^{2} + 2 x + 3} = \frac{3 (10 x + 2)}{10 (5 x^{2} + 2 x + 3)} + \frac{7}{5 (5 x^{2} + 2 x + 3)}$ $\int \frac{3 x + 2}{5 x^{2} + 2 x + 3} = \int \frac{3 (10 x + 2)}{10 (5 x^{2} + 2 x + 3)} + \int \frac{7}{5 (5 x^{2} + 2 x + 3)}$ $\frac{3}{10} \int \frac{10 x + 2}{5 x^{2} + 2 x + 3} d x + \frac{7}{5} \int \frac{1}{5 x^{2} + 2 x + 3} d x$ $u = 5 x^{2} + 2 x + 3 \to d u = (10 x + 2) d x$ $\frac{3}{10} \int \frac{d u}{u} = \frac{3}{10} \ln (u) = \frac{3}{10} \ln (5 x^{2} + 2 x + 3) (A)$ $and \frac{7}{5} \int \frac{1}{5 x^{2} + 2 x + 3} d x = \frac{7}{5} \int \frac{1}{{(\sqrt{5} x + \frac{1}{\sqrt{5}})}^{2} + \frac{14}{5}} d x$ $\sqrt{5} x + \frac{1}{\sqrt{5}} = u \to d u = \sqrt{5} d x$ $\frac{7}{5 \sqrt{5}} \int \frac{5}{14 (\frac{5 u^{2}}{14} + 1)} d u = \frac{1}{2 \sqrt{5}} \int \frac{1}{\frac{5 u^{2}}{14} + 1} d u$ $p = \sqrt{\frac{5}{14}} v \to d p = \sqrt{\frac{5}{14}} d v$ $\frac{1}{5} \sqrt{\frac{7}{2}} \int \frac{1}{p^{2} + 1} d p = \frac{1}{5} \sqrt{\frac{7}{2}} \tan^{- 1} (p)$ $= \frac{1}{5} \sqrt{\frac{7}{2}} \tan^{- 1} (\sqrt{\frac{5}{14}} v) = \frac{1}{5} \sqrt{\frac{7}{2}} \tan^{- 1} (\frac{5 x + 1}{\sqrt{14}}) (B)$ $∴ Solution = A + B$ $y (x) = \frac{3}{10} \ln (5 x^{2} + 2 x + 3) + \frac{\sqrt{14}}{10} \tan^{- 1} (\frac{5 x + 1}{\sqrt{14}})$ $H oly . . . . . shit . . . . what a terrible differantial equation$ $XD$
	Commented by Frix last updated on 31/Oct/24

	$Why terrible ? Try this :$ $\int \sin x^{3} d x = ?$
	Answered by MrGaster last updated on 30/Oct/24

	$= \frac{1}{5} \int \frac{15 x + 10}{5 x^{2} + 2 x + 3} d x$ $= \frac{1}{5} \int \frac{15 x + 2 + 8}{5 x^{2} + 2 x + 3} d x$ $= \frac{1}{5} (\int \frac{15 x + 2}{5 x^{2} + 2 x + 3} d x + \int \frac{8}{5 x^{2} + 2 x + 3} d x)$ $= \frac{1}{5} (\int \frac{d (5 x^{2} + 2 x + 3)}{5 x^{2} + 2 x + 3} + \int \frac{8}{5 (x^{2} + \frac{2}{5} x + \frac{3}{5}} d x)$ $= \frac{1}{5} (\ln ∣ 5 x^{2} + 2 x + 3 ∣ + 8 \int \frac{1}{5 (x^{2} + \frac{2}{5} x + \frac{3}{5})} d x)$ $= \frac{1}{5} (\ln ∣ 5 x^{2} + 2 x + 3 ∣ + \frac{8}{5} \int \frac{1}{{(x + \frac{1}{5})}^{2} + {(\frac{\sqrt{14}}{5})}^{2}} d x)$ $= \frac{1}{5} \ln ∣ 5 x^{2} + 2 x + 3 ∣ + \frac{8}{25} \cdot \frac{5}{\sqrt{14}} \arctan (\frac{x + \frac{1}{5}}{\frac{\sqrt{14}}{5}}) + C$ $= \frac{1}{5} \ln ∣ 5 x^{2} + 2 x + 3 ∣ + \frac{8}{\sqrt{14}} \arctan (\frac{5 x + 1}{\sqrt{14}}) + C$
	Answered by Ghisom last updated on 30/Oct/24

	$\int \frac{3 x + 2}{5 x^{2} + 2 x + 3} d x =$ $[t = \frac{5 x + 1}{\sqrt{14}}]$ $= \frac{3}{5} \int \frac{t}{t^{2} + 1} d t + \frac{\sqrt{14}}{10} \int \frac{d t}{t^{2} + 1} =$ $= \frac{3}{10} \ln (t^{2} + 1) + \frac{\sqrt{14}}{10} \arctan t =$ $= \frac{3}{10} \ln (5 x^{2} + 2 x + 3) + \frac{\sqrt{14}}{10} \arctan \frac{5 x + 1}{\sqrt{14}} + C$
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