What is the integral from 0 to 3 of (2t)/((t-4)^2) ?

The integral from 0 to 3 of (2t)/((t-4)^2) is 2(-2ln(2)+3)

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integral from 0 to 3 of (2t)/((t-4)^2)

\int_{0}^{3} \frac{2 t}{( t - 4 ) ^{2}} d t

2 (- 2 ln (2) + 3)

Decimal Notation

3.22741 \dots

Solution steps

\int_{0}^{3} \frac{2 t}{( t - 4 ) ^{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}^{3} \frac{t}{( t - 4 ) ^{2}} d t

Apply u-substitution

: \int_{- 4}^{- 1} \frac{u + 4}{u ^{2}} d u

= 2 \cdot \int_{- 4}^{- 1} \frac{u + 4}{u ^{2}} d u

Expand

\frac{u + 4}{u ^{2}} : \frac{1}{u} + \frac{4}{u ^{2}}

= 2 \cdot \int_{- 4}^{- 1} \frac{1}{u} + \frac{4}{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_{- 4}^{- 1} \frac{1}{u} d u + \int_{- 4}^{- 1} \frac{4}{u ^{2}} d u)

\int_{- 4}^{- 1} \frac{1}{u} d u = - 2 ln (2)

\int_{- 4}^{- 1} \frac{4}{u ^{2}} d u = 3

= 2 (- 2 ln (2) + 3)

\frac{x ^{2} - 3 x + 2}{- x ^{2} - 4 x + 4}

\frac{\partial}{\partial x} (y^{4} sin (5 x))

\frac{1}{s ^{2} + 9}

\frac{\partial}{\partial y} (\frac{y}{x + y})

What is the integral from 0 to 3 of (2t)/((t-4)^2) ?
The integral from 0 to 3 of (2t)/((t-4)^2) is 2(-2ln(2)+3)