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

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

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

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

4 (- 2 ln (2) + 3)

Decimal Notation

6.45482 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Expand

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

= 4 \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

= 4 (\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

= 4 (- 2 ln (2) + 3)

\int \frac{1}{x} sin (x) d x

\int_{1}^{\infty} \frac{1}{x ^{1}} d x

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

\frac{d}{d x} (8 x e^{x} - 8 e^{x})

y = 4 x - 3 x^{2}, (2, - 4)

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