Find the integral: ∫ (x+2)/(2x^2+6x+5) dx

The correct answer is Option (4) → $\frac{1}{4} \ln|2x^2 + 6x + 5| + \frac{1}{2} \tan^{-1}(2x + 3) + C$

Using the formula, we express

$x + 2 = A \frac{d}{dx}(2x^2 + 6x + 5) + B = A(4x + 6) + B$

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, $\int \frac{x+2}{2x^2+6x+5} dx = \frac{1}{4} \int \frac{4x+6}{2x^2+6x+5} dx + \frac{1}{2} \int \frac{dx}{2x^2+6x+5}$

$= \frac{1}{4} I_1 + \frac{1}{2} I_2 \text{ (say)} \dots (1)$

In $I_1$, put $2x^2 + 6x + 5 = t$, so that $(4x + 6) \, dx = dt$

Therefore,

$I_1 = \int \frac{dt}{t} = \log |t| + C_1$

$= \log |2x^2 + 6x + 5| + C_1 \dots (2)$

and

$I_2 = \int \frac{dx}{2x^2 + 6x + 5} = \frac{1}{2} \int \frac{dx}{x^2 + 3x + \frac{5}{2}}$

$= \frac{1}{2} \int \frac{dx}{\left( x + \frac{3}{2} \right)^2 + \left( \frac{1}{2} \right)^2}$

Put $x + \frac{3}{2} = t$, so that $dx = dt$, we get

$I_2 = \frac{1}{2} \int \frac{dt}{t^2 + \left( \frac{1}{2} \right)^2} = \frac{1}{2 \times \frac{1}{2}} \tan^{-1} 2t + C_2$

$= \tan^{-1} 2\left( x + \frac{3}{2} \right) + C_2 = \tan^{-1} (2x + 3) + C_2 \dots (3)$

Using (2) and (3) in (1), we get

$\int \frac{x+2}{2x^2 + 6x + 5} \, dx = \frac{1}{4} \log |2x^2 + 6x + 5| + \frac{1}{2} \tan^{-1} (2x + 3) + C$

where, $C = \frac{C_1}{4} + \frac{C_2}{2}$