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

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

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

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

2 (- ln (3) + 2)

Decimal Notation

1.80277 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

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

= 2 \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

= 2 (\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

= 2 (- ln (3) + 2)

\frac{\partial}{\partial x} (\frac{9 x - 2 y}{9 x + 2 y})

f (x) = \frac{x ^{\frac{1}{3}} - 4}{x ^{\frac{7}{2}}}

c \cdot sin (\frac{π x}{L})

\frac{3}{4} s^{2}

\frac{\partial}{\partial x} (\frac{x + 2 y}{x ^{2} - y})

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