Evaluate $\int \frac{3x-1}{\sqrt{x^2+9}} dx$. |
$3\sqrt{x^2 + 9} - \ln|x + \sqrt{x^2 + 9}| + C$ $3\sqrt{x^2 + 9} + \ln|x + \sqrt{x^2 + 9}| + C$ $\frac{3}{2}\sqrt{x^2 + 9} - \ln|x + \sqrt{x^2 + 9}| + C$ $3\sqrt{x^2 + 9} - \arctan\left(\frac{x}{3}\right) + C$ |
$3\sqrt{x^2 + 9} - \ln|x + \sqrt{x^2 + 9}| + C$ |
The correct answer is Option (1) → $3\sqrt{x^2 + 9} - \ln|x + \sqrt{x^2 + 9}| + C$ Let $I = \int \frac{3x - 1}{\sqrt{x^2 + 9}} dx$ $\Rightarrow I = \int \frac{3x}{\sqrt{x^2 + 9}} dx - \int \frac{1}{\sqrt{x^2 + 9}} dx$ Let $I = I_1 - I_2$ Now, $I_1 = \int \frac{3x}{\sqrt{x^2 + 9}} dx$ Put $x^2 + 9 = t^2 \Rightarrow 2x \, dx = 2t \, dt \Rightarrow x \, dx = t \, dt$ $∴I_1 = 3 \int \frac{t}{t} dt$ $= 3 \int dt = 3t + C_1 = 3\sqrt{x^2 + 9} + C_1$ and $I_2 = \int \frac{1}{\sqrt{x^2 + 9}} dx = \int \frac{1}{\sqrt{x^2 + (3)^2}} dx$ $= \log |x + \sqrt{x^2 + 9}| + C_2$ $∴I = 3\sqrt{x^2 + 9} + C_1 - \log |x + \sqrt{x^2 + 9}| - C_2$ $= 3\sqrt{x^2 + 9} - \log |x + \sqrt{x^2 + 9}| + C \quad [∵C = C_1 - C_2]$ |