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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

5 (- ln (5) + 4)

Decimal Notation

11.95281 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Expand

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

= 5 \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

= 5 (\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

= 5 (- ln (5) + 4)

\int \frac{2}{( x - 1 ) ^{3}} d x

x \to 0 + lim (\frac{1}{e ^{x} - 1} - \frac{1}{x})

y = 4 - x^{2}, y = 0

cos (3 t) - cos (7 t)

x^{3} + 3

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