What is the integral from 0 to 1 of x/((2-x)^2) ?

The integral from 0 to 1 of x/((2-x)^2) is 1-ln(2)

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integral from 0 to 1 of x/((2-x)^2)

\int_{0}^{1} \frac{x}{( 2 - x ) ^{2}} d x

1 - ln (2)

Decimal Notation

0.30685 \dots

Solution steps

\int_{0}^{1} \frac{x}{( 2 - x ) ^{2}} d x

Apply Integration By Parts

: [\frac{x}{2 - x} - \int \frac{1}{2 - x} d x]_{0}^{1}

= [\frac{x}{2 - x} - \int \frac{1}{2 - x} d x]_{0}^{1}

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

= [\frac{x}{2 - x} - (- ln ∣ 2 - x ∣)]_{0}^{1}

Simplify

= [\frac{x}{2 - x} + ln ∣ 2 - x ∣]_{0}^{1}

Compute the boundaries:

1 - ln (2)

= 1 - ln (2)

\frac{d y}{d x} = x (y - 1)

\frac{d}{d x} (4 sec (x) + tan (x))

x \to 0 lim (\frac{x}{e ^{5 x} - 1})

\frac{d}{d x} (x arctan (x))

\int x^{\frac{7}{3}} d x

What is the integral from 0 to 1 of x/((2-x)^2) ?
The integral from 0 to 1 of x/((2-x)^2) is 1-ln(2)