Find the integral: x+2 2 x^ 2 +6 x+5 d x

Question

Find the integral: $\int \frac{x+2}{2 x^{2}+6 x+5} d x$

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Answer

Let,$ I = ∫\frac{(x+2)}{(2x^2+6x+5)}dx$
we express
$x+2=\mathrm{A} \frac{d}{d x}\left(2 x^{2}+6 x+5\right)+\mathrm{B} = A(4x + 6) + B$
Equating the coefficients of x and the constant terms from both sides, we get
$4A = 1$ and $6A + B = 2$ or $A=\frac{1}{4}$ and $B=\frac{1}{2}$
Therefore, $\int \frac{x+2}{2 x^{2}+6 x+5}=\frac{1}{4} \int \frac{4 x+6}{2 x^{2}+6 x+5} d x+\frac{1}{2} \int \frac{d x}{2 x^{2}+6 x+5}$
$= \frac{1}{4} \mathrm{I}_{1}+\frac{1}{2} \mathrm{I}_{2}$ .....(i)
In $I_1,$ put $2x^2 + 6x + 5 = t,$ so that $(4x + 6) dx = dt$
Therefore, $I_{1}=\int \frac{d t}{t}=\log |t|+C_{1}$
$= \log |2x^2 + 6x + 5| + C .....(ii)$
and $I_{2}=\int \frac{d x}{2 x^{2}+6 x+5}=\frac{1}{2} \int \frac{d x}{x^{2}+3 x+\frac{5}{2}}$
$= \frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}$
Put $x+\frac{3}{2}=t$, so that $dx = dt,$ we get
$I_{2}=\frac{1}{2} \int \frac{d t}{t^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{1}{2 \times \frac{1}{2}} \tan ^{-1} 2 t+C_{2}$
$\tan ^{-1} 2\left(x+\frac{3}{2}\right)+\mathrm{C}_{2}=\tan ^{-1}(2 x+3)+\mathrm{C}_{2}......(iii)$
Using (ii) and (iii) in (i), we get
$\int \frac{x+2}{2 x^{2}+6 x+5} d x=\frac{1}{4} \log \left|2 x^{2}+6 x+5\right|+\frac{1}{2} \tan ^{-1}(2 x+3)+\mathrm{C}$
where, $C=\frac{C_{1}}{4}+\frac{C_{2}}{2}$