Practicing Success
Practicing Success
CUET Preparation Today
The integral $\int e^x\left(\log x+\frac{1}{x^2}\right) d x$ equals for some arbitrary constant k |
$e^x\left(\log x+\frac{1}{x^2}\right)+k$ $e^x\left(\log x-\frac{1}{x}\right)+k$ $e^x\left(\log x-\frac{1}{x^2}\right)+k$ $e^x\left(\log x+\frac{1}{x}\right)+k$ |
$e^x\left(\log x-\frac{1}{x}\right)+k$ |
The correct answer is Option (2) → $e^x\left(\log x-\frac{1}{x}\right)+k$ |
