Integrate the function 5 x-2 1+2 x+3 x^ 2

Question

Integrate the function $\frac{5 x-2}{1+2 x+3 x^{2}}$

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Answer

Given integral is: $\frac{5 x-2}{1+2 x+3 x^{2}}$
Let 5x - 2 = $A \frac{d}{d x}\left(1+2 x+3 x^{2}\right)+B$
$\Rightarrow$ 5x - 2 = A(2 + 6x) + B
Now, equating the coefficients of x and constant term on both sides, we get,
6A = 5 $\Rightarrow A=\frac{5}{6}$
2A + B = -2 $\Rightarrow B=-\frac{11}{3}$
$\Rightarrow 5 x-2=\frac{5}{6}(2+6 x)+\left(-\frac{11}{3}\right)$
$\therefore~ \int \frac{5 x+1}{1+2 x+3 x^{2}} d x=\int \frac{\frac{5}{6}(2+6 x)-\frac{11}{3}}{1+2 x+3 x^{2}} d x$
$= \frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^{2}} d x-\frac{11}{3} \int \frac{1}{1+2 x+3 x^{2}} d x$
Now, in $\int \frac{2+6 x}{1+2 x+3 x^{2}} d x$
Let 1 + 2x + 3x² = t
$\Rightarrow$ (2 + 6x)dx = dt
$\therefore~\int \frac{2+6 \mathrm{x}}{1+2 \mathrm{x}+3 \mathrm{x}^{2}} \mathrm{dx}=\int \frac{\mathrm{dt}}{\mathrm{t}}=\log |\mathrm{t}|$
= log|1 + 2x + 3x²| ....…(1)
Also in $\int \frac{1}{1+2 x+3 x^{2}} d x$
$1+2 x+3 x^{2}=1+3\left(x^{2}+\frac{2}{3} x\right)$
$=3\left[\left(\mathrm{x}+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}\right]$
$\Rightarrow \int \frac{1}{1+2 x+3 x^{2}} d x=\frac{1}{3} \int \frac{1}{\left[\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}\right]} d x$
$= \frac{1}{3}\left[\frac{1}{\frac{\sqrt{2}}{3}} \tan ^{-1}\left(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right)\right]$
$= \frac{1}{3}\left[\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 \mathrm{x}+1}{\sqrt{2}}\right)\right]$
$=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)$ ......(ii)
Thus, from (1) and (2), we get,
$\Rightarrow \int \frac{5 x+1}{1+2 x+3 x^{2}} d x=\frac{5}{6}\left[\log \left|1+2 x+3 x^{2}\right|\right]-\frac{11}{3}\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)\right]+C$
$= \frac{5}{6}\left[\log \left|1+2 \mathrm{x}+3 \mathrm{x}^{2}\right|\right]-\frac{11}{3 \sqrt{2}} \tan ^{-1}\left(\frac{3 \mathrm{x}+1}{\sqrt{2}}\right)+\mathrm{C}$

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