What is the integral of ln(1-sqrt(x)) ?

The integral of ln(1-sqrt(x)) is xln(1-sqrt(x))-1/2 x-sqrt(x)-ln|-sqrt(x)+1|-1/2+2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ln(1-sqrt(x))

\int ln (1 - x) d x

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

Solution steps

\int ln (1 - x) d x

Apply Integration By Parts

: x ln (1 - x) - \int - \frac{x}{2 ( 1 - x )} d x

= x ln (1 - x) - \int - \frac{x}{2 ( 1 - x )} d x

\int - \frac{x}{2 ( 1 - x )} d x = \frac{1}{2} (1 - x)^{2} - 2 (1 - x) + ln 1 - x

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

Simplify

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

= x ln (1 - x) - \frac{1}{2} x - x - ln - x + 1 - \frac{1}{2} + 2

Add a constant to the solution

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

y = arccot (2 t)

\int \frac{4 x ^{2}}{( 16 + x ^{2} ) ^{2}} d x

y^{^{''}} - 6 y^{^{'}} - 7 y = 15 e^{2 t}

\int - \frac{5}{x ^{2}} d x

What is the integral of ln(1-sqrt(x)) ?
The integral of ln(1-sqrt(x)) is xln(1-sqrt(x))-1/2 x-sqrt(x)-ln|-sqrt(x)+1|-1/2+2+C