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We want to express $3x+5$ in the form $A(2x+2) + B$, where $2x+2$ is the derivative of $x^2+2x+4$. $$3x+5 = A(2x+2) + B$$ $$3x+5 = 2Ax + 2A + B$$ Comparing coefficients, we have: $2A = 3 \implies A = \frac{3}{2}$ $2A + B = 5 \implies 2(\frac{3}{2}) + B = 5 \implies 3 + B = 5 \implies B = 2$ So, $3x+5 = \frac{3}{2}(2x+2) + 2$
Now we can rewrite the integral as: $$\int\frac{3x+5}{\sqrt{x^{2}+2x+4}}dx = \int\frac{\frac{3}{2}(2x+2) + 2}{\sqrt{x^{2}+2x+4}}dx$$ $$= \frac{3}{2}\int\frac{2x+2}{\sqrt{x^{2}+2x+4}}dx + 2\int\frac{1}{\sqrt{x^{2}+2x+4}}dx$$
Let $u = x^2+2x+4$, then $du = (2x+2)dx$. $$\frac{3}{2}\int\frac{2x+2}{\sqrt{x^{2}+2x+4}}dx = \frac{3}{2}\int\frac{1}{\sqrt{u}}du = \frac{3}{2}\int u^{-\frac{1}{2}}du$$ $$= \frac{3}{2} \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C_1 = \frac{3}{2} \cdot 2\sqrt{u} + C_1 = 3\sqrt{x^2+2x+4} + C_1$$
$$2\int\frac{1}{\sqrt{x^{2}+2x+4}}dx = 2\int\frac{1}{\sqrt{(x+1)^2 + 3}}dx$$ Let $x+1 = \sqrt{3}\tan\theta$, then $dx = \sqrt{3}\sec^2\theta d\theta$. $$2\int\frac{1}{\sqrt{(x+1)^2 + 3}}dx = 2\int\frac{\sqrt{3}\sec^2\theta}{\sqrt{3\tan^2\theta + 3}}d\theta = 2\int\frac{\sqrt{3}\sec^2\theta}{\sqrt{3}\sec\theta}d\theta$$ $$= 2\int\sec\theta d\theta = 2\ln|\sec\theta + \tan\theta| + C_2$$ Since $x+1 = \sqrt{3}\tan\theta$, $\tan\theta = \frac{x+1}{\sqrt{3}}$. $\sec\theta = \sqrt{1 + \tan^2\theta} = \sqrt{1 + \frac{(x+1)^2}{3}} = \sqrt{\frac{3 + x^2 + 2x + 1}{3}} = \sqrt{\frac{x^2+2x+4}{3}}$ $$2\ln|\sec\theta + \tan\theta| + C_2 = 2\ln|\sqrt{\frac{x^2+2x+4}{3}} + \frac{x+1}{\sqrt{3}}| + C_2$$ $$= 2\ln|\frac{\sqrt{x^2+2x+4} + x+1}{\sqrt{3}}| + C_2 = 2\ln|\sqrt{x^2+2x+4} + x+1| - 2\ln\sqrt{3} + C_2$$ $$= 2\ln|\sqrt{x^2+2x+4} + x+1| + C_3$$
$$\int\frac{3x+5}{\sqrt{x^{2}+2x+4}}dx = 3\sqrt{x^2+2x+4} + 2\ln|\sqrt{x^2+2x+4} + x+1| + C$$
Final Answer: $3\sqrt{x^2+2x+4} + 2\ln|\sqrt{x^2+2x+4} + x+1| + C$
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