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

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

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

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

1 - ln (2)

Decimal

0.30685 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

1 - ln (2)

= 1 - ln (2)

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\frac{\partial}{\partial x} (e^{i x} cos (i y))

\int ln (x) \cdot \frac{1}{x} d x

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