The integral $\int \log _ { \frac { 1 } { 2 } } x d x$ is

A) $- \frac { x ( 2 \ln x - 1 ) } { \ln 2 } + C$
B) $- \frac { x ( 2 \ln x + 1 ) } { \ln 2 } + C$
C) $\frac { x ( \ln x - 1 ) } { \ln 2 } + C$
D) $- \frac { x ( \ln x - 1 ) } { \ln 2 } + \mathrm { C }$
E) $- \frac { x ( \ln x + 1 ) } { \ln 2 } + C$

Question

The integral $\int \log _ { \frac { 1 } { 2 } } x d x$ is

A) $- \frac { x ( 2 \ln x - 1 ) } { \ln 2 } + C$
B) $- \frac { x ( 2 \ln x + 1 ) } { \ln 2 } + C$
C) $\frac { x ( \ln x - 1 ) } { \ln 2 } + C$
D) $- \frac { x ( \ln x - 1 ) } { \ln 2 } + \mathrm { C }$
E) $- \frac { x ( \ln x + 1 ) } { \ln 2 } + C$

Quizplus · Accepted Answer

The integral $\int \log _ { \frac { 1 } { 2 } } x d x$ can be solved by recognizing that $\log _ { \frac { 1 } { 2 } } x = \frac{\ln x}{\ln \frac{1}{2}} = -\frac{\ln x}{\ln 2}$. Using integration by parts, where $u = \ln x$ and $dv = dx$, leads to the result $- \frac { x ( \ln x - 1 ) } { \ln 2 } + C$, which matches choice D.

The Integral ∫log⁡12xdx\int \log _ { \frac { 1 } { 2 } } x d x∫log21​​xdx

The Integral $\int \log _ { \frac { 1 } { 2 } } x d x$