What is the integral from 0 to 2 of t/((3-t)^2) ?

The integral from 0 to 2 of t/((3-t)^2) is 2-ln(3)

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integral from 0 to 2 of t/((3-t)^2)

\int_{0}^{2} \frac{t}{( 3 - t ) ^{2}} d t

2 - ln (3)

Decimal

0.90138 \dots

Solution steps

\int_{0}^{2} \frac{t}{( 3 - t ) ^{2}} d t

Apply Integration By Parts

: [\frac{t}{3 - t} - \int \frac{1}{3 - t} d t]_{0}^{2}

= [\frac{t}{3 - t} - \int \frac{1}{3 - t} d t]_{0}^{2}

\int \frac{1}{3 - t} d t = - ln ∣ 3 - t ∣

= [\frac{t}{3 - t} - (- ln ∣ 3 - t ∣)]_{0}^{2}

Simplify

= [\frac{t}{3 - t} + ln ∣ 3 - t ∣]_{0}^{2}

Compute the boundaries:

2 - ln (3)

= 2 - ln (3)

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What is the integral from 0 to 2 of t/((3-t)^2) ?
The integral from 0 to 2 of t/((3-t)^2) is 2-ln(3)