Иант ํo 15 \[ \int \frac{3 x+5}{\sqrt{x^{2}+4 x+5}} d x \]

Step 1 ：Let \(u = x^2 + 4x + 5\), then \(\frac{du}{dx} = 2x + 4\)

Step 2 ：Separate the integral into two parts: \(\int \frac{3 x+5}{\sqrt{x^{2}+4 x+5}} d x = \int \frac{2x + 4}{\sqrt{x^{2}+4 x+5}} d x + \int \frac{x + 1}{\sqrt{x^{2}+4 x+5}} d x\)

Step 3 ：Now substitute u: \(\int \frac{2x + 4}{\sqrt{x^{2}+4 x+5}} d x = \frac{1}{2} \int \frac{du}{\sqrt{u}}\) and \(\int \frac{x + 1}{\sqrt{x^{2}+4 x+5}} d x = \int \frac{\frac{du}{2} - 1}{\sqrt{u}}\)

Step 4 ：Solve the integrals: \(\frac{1}{2} \int \frac{du}{\sqrt{u}} = \sqrt{u}\) and \(\int \frac{\frac{du}{2} - 1}{\sqrt{u}} = \frac{1}{2}u\sqrt{u} - \frac{9}{2}\ln(u)\)

Step 5 ：Re-substitute x: \(x\sqrt{x^{2}+4x+5} + \frac{9}{2}\ln(x^2+4x+5) + C\)