∫ x x ((1/ x )+ ln 2 x + ln x ) dx is equal to (where C is contant of integration)

Question

∫ x x ((1/ x )+ ln 2 x + ln x ) dx is equal to (where C is contant of integration) (A) x x ( ln 2 x -(1/ x ))+ C (B) x x ( ln x - x )+ C (C) (xx ln 2 x/2)+C (D) x x ln x + C

Tardigrade Admin · Accepted Answer

Use ∫(f(x)+x f prime(x)) d x=x f(x)+C

Q. $\int x^{x} (\frac{1}{x} + ln^{2} x + ln x) d x$ is equal to (where $C$ is contant of integration)

$x^{x} (ln^{2} x - \frac{1}{x}) + C$

$x^{x} (ln x - x) + C$

$\frac{x ^{x} l n ^{2} x}{2} + C$

$x^{x} ln x + C$

Use $\int (f (x) + x f^{'} (x)) d x = x f (x) + C$

Q. ∫xx(x1​+ln2x+lnx)dx is equal to (where C is contant of integration)

xx(ln2x−x1​)+C

xx(lnx−x)+C

2xxln2x​+C

xxlnx+C