Question

∫ log x · (log x + 2) dx = ?

Options

• x (log x)² + c

• x log x + c

• e^x (log x)² + c

• (log x)² + c

Answer 1

x (log x)² + c

Explanation:

We have,

I = ∫ log x · (log x + 2) dx

I = ∫ (log x)² dx + ∫ 2 log x dx

I = `(log x)^2 int "dx" - int (("d"(log x)^2)/"dx" int "dx") "dx" + int 2 log x "d"x + "C"`

I = `x (log x)^2 - int (2 log x)/x * x  "dx" + int 2 log x "dx" + "C"`

I = x (log x)² - ∫ 2 log x dx + ∫ 2 log x dx + C

I = (x log x)² + C