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

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

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

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

2 (- ln (5) + 4)

Decimal

4.78112 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

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

= 2 \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

= 2 (\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

= 2 (- ln (5) + 4)

x \to 0 lim (ln (1 + x))

\frac{\partial}{\partial z} (8 x^{\frac{y}{z}})

3 (t - 2)^{2} + 13 (t - 2) + 17

n = 0 \sum \infty n (1 - p)^{n - 1}

\int_{1}^{3} \frac{1}{x} d x

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