Solve the integral: I = ∫ (3x+5)/(x^2+4x+7) dx

The correct answer is Option (2) → $\frac{3}{2} \ln(x^2 + 4x + 7) - \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x + 2}{\sqrt{3}}\right) + C$

Rewrite the numerator as:

$3x+5 = A \frac{d}{dx}(x^2+4x+7) + B$

$⇒3x+5 = A(2x+4) + B$

Find the value of A and B by comparing the coefficients of like terms as:

$2A = 3 ⇒A = \frac{3}{2}$

$4A + B = 5 ⇒B = -1$

Substitute the values of A and B in the given integral and integrates the same as:

$I = \int \frac{\frac{3}{2}(2x+4)-1}{x^2+4x+7} dx$

$⇒I = \frac{3}{2} \int \frac{2x+4}{x^2+4x+7} dx - \int \frac{1}{(x+2)^2 + (\sqrt{3})^2} dx$

$⇒I = \frac{3}{2} \log |x^2+4x+7| - \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x+2}{\sqrt{3}} \right) + C$

Where, C is the constant of integration.