What is the integral from 0 to 4 of (8t)/((t-5)^2) ?

The integral from 0 to 4 of (8t)/((t-5)^2) is 8(-ln(5)+4)

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integral from 0 to 4 of (8t)/((t-5)^2)

\int_{0}^{4} \frac{8 t}{( t - 5 ) ^{2}} d t

8 (- ln (5) + 4)

Solution steps

\int_{0}^{4} \frac{8 t}{( t - 5 ) ^{2}} d t

Take the constant out:

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

= 8 \cdot \int_{0}^{4} \frac{t}{( t - 5 ) ^{2}} d t

Apply u-substitution

: \int_{- 5}^{- 1} \frac{u + 5}{u ^{2}} d u

= 8 \cdot \int_{- 5}^{- 1} \frac{u + 5}{u ^{2}} d u

Expand

\frac{u + 5}{u ^{2}} : \frac{1}{u} + \frac{5}{u ^{2}}

= 8 \cdot \int_{- 5}^{- 1} \frac{1}{u} + \frac{5}{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

= 8 (\int_{- 5}^{- 1} \frac{1}{u} d u + \int_{- 5}^{- 1} \frac{5}{u ^{2}} d u)

\int_{- 5}^{- 1} \frac{1}{u} d u = - ln (5)

\int_{- 5}^{- 1} \frac{5}{u ^{2}} d u = 4

= 8 (- ln (5) + 4)

x \to 0 lim (\frac{∣ x - 0 ∣}{x - 0})

\frac{\partial}{\partial x} (7 x^{2} + y^{2})

f (x) = x^{4} - 5 x^{2} + 2, at x = 2

\frac{d}{d x} (5 (e^{x} cos (x) - e^{x} sin (x)))

What is the integral from 0 to 4 of (8t)/((t-5)^2) ?
The integral from 0 to 4 of (8t)/((t-5)^2) is 8(-ln(5)+4)