Find: $\int \frac{3x+5}{\sqrt{x^2+2x+4}} dx$. |
$3\sqrt{x^2+2x+4} + 2\ln|x+1 + \sqrt{x^2+2x+4}| + C$ $\frac{3}{2}\sqrt{x^2+2x+4} + 2\ln|x+1 + \sqrt{x^2+2x+4}| + C$ $3\sqrt{x^2+2x+4} - 2\ln|x+1 + \sqrt{x^2+2x+4}| + C$ $2\sqrt{x^2+2x+4} + 3\ln|x+1 + \sqrt{x^2+2x+4}| + C$ |
$3\sqrt{x^2+2x+4} + 2\ln|x+1 + \sqrt{x^2+2x+4}| + C$ |
The correct answer is Option (1) → $3\sqrt{x^2+2x+4} + 2\ln|x+1 + \sqrt{x^2+2x+4}| + C$ Let $I = \int \frac{3x+5}{\sqrt{x^2+2x+4}} dx$ $= \int \left[ \frac{3(x+1)}{\sqrt{x^2+2x+4}} + \frac{2}{\sqrt{(x+1)^2+(\sqrt{3})^2}} \right] dx$ $I = I_1 + I_2$ where, $I_1 = 3 \int \frac{x+1}{\sqrt{x^2+2x+4}} dx$ Let $x^2+2x+4 = t$ $(2x+2) dx = dt$ $(x+1) dx = \frac{dt}{2}$ $I_1 = \frac{3}{2} \int t^{-1/2} dt$ $I_1= \frac{3}{2} \frac{t^{1/2}}{1/2} + C_1$ $∴I_1= 3\sqrt{x^2+2x+4} + C_1$ and $I_2 = 2 \int \frac{1}{\sqrt{(x+1)^2+(\sqrt{3})^2}} dx$ $I_2 = 2 \log |x+1 + \sqrt{x^2+2x+4}| + C_2$ Thus, $I = 3\sqrt{x^2+2x+4} + 2\log |x+1 + \sqrt{x^2+2x+4}| + C$ where $C = C_1 + C_2$ |