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

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

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

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

2 (1 - ln (2))

Decimal

0.61370 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \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}

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

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

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

Simplify

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

Compute the boundaries:

1 - ln (2)

= 2 (1 - ln (2))

\frac{11}{ln ( x )}

x \to 2 lim (ln (x^{2} (3 - x)))

4 x + 12, x^{2}, 8 - x^{2}

4^{2 x^{2}}

2 + 3 x

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