What is the integral from 0 to 2 of (5t)/((t-3)^2) ?

The integral from 0 to 2 of (5t)/((t-3)^2) is 5(-ln(3)+2)

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integral from 0 to 2 of (5t)/((t-3)^2)

\int_{0}^{2} \frac{5 t}{( t - 3 ) ^{2}} d t

5 (- ln (3) + 2)

Decimal Notation

4.50693 \dots

Solution steps

\int_{0}^{2} \frac{5 t}{( t - 3 ) ^{2}} d t

Take the constant out:

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

= 5 \cdot \int_{0}^{2} \frac{t}{( t - 3 ) ^{2}} d t

Apply u-substitution

: \int_{- 3}^{- 1} \frac{u + 3}{u ^{2}} d u

= 5 \cdot \int_{- 3}^{- 1} \frac{u + 3}{u ^{2}} d u

Expand

\frac{u + 3}{u ^{2}} : \frac{1}{u} + \frac{3}{u ^{2}}

= 5 \cdot \int_{- 3}^{- 1} \frac{1}{u} + \frac{3}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 5 (\int_{- 3}^{- 1} \frac{1}{u} d u + \int_{- 3}^{- 1} \frac{3}{u ^{2}} d u)

\int_{- 3}^{- 1} \frac{1}{u} d u = - ln (3)

\int_{- 3}^{- 1} \frac{3}{u ^{2}} d u = 2

= 5 (- ln (3) + 2)

\frac{d}{d x} (ln (- 5 x))

\frac{\frac{1}{3}}{s}

t \cdot \frac{d y}{d t} = t^{3} + 30 t^{3} y

n = 0 \sum \infty n! (x - 3)^{n}

f (x) = 2 x^{\frac{1}{3}} + \frac{x}{3} - \frac{1}{4} x^{3}

What is the integral from 0 to 2 of (5t)/((t-3)^2) ?
The integral from 0 to 2 of (5t)/((t-3)^2) is 5(-ln(3)+2)