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

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

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

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

3 (2 - ln (3))

Decimal

2.70416 \dots

Solution steps

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

Take the constant out:

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

= 3 \cdot \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}

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

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

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

Simplify

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

Compute the boundaries:

2 - ln (3)

= 3 (2 - ln (3))

y = 3 x^{2} + 5 x, (- 3, - 13)

x \to 3 lim (e^{x})

x \to \infty lim (tan (x))

\frac{d}{d x} (ln (x^{2} + 3))

\frac{d}{d x} (3^{(x^{4} + 1)^{3}})

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