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

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

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

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

2 (1 - ln (2))

Decimal Notation

0.61370 \dots

Solution steps

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

Take the constant out:

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

= 2 \cdot \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}

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

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

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

Simplify

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

Compute the boundaries:

1 - ln (2)

= 2 (1 - ln (2))

\int \frac{x ^{3}}{x ^{2} - 9} d x

\frac{9}{( 1 - x ) ^{2}}

n = 1 \sum \infty \frac{1}{n ^{5}}

f (x) = (12 - x) 24 x - 144

y^{^{'''}} - y^{^{''}} + 6 y^{^{'}} - 6 y = 0

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