What is the integral from 0 to 5 of (5t)/((t-6)^2) ?

The integral from 0 to 5 of (5t)/((t-6)^2) is 5(-ln(6)+5)

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integral from 0 to 5 of (5t)/((t-6)^2)

\int_{0}^{5} \frac{5 t}{( t - 6 ) ^{2}} d t

5 (- ln (6) + 5)

Decimal Notation

16.04120 \dots

Solution steps

\int_{0}^{5} \frac{5 t}{( t - 6 ) ^{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}^{5} \frac{t}{( t - 6 ) ^{2}} d t

Apply u-substitution

: \int_{- 6}^{- 1} \frac{u + 6}{u ^{2}} d u

= 5 \cdot \int_{- 6}^{- 1} \frac{u + 6}{u ^{2}} d u

Expand

\frac{u + 6}{u ^{2}} : \frac{1}{u} + \frac{6}{u ^{2}}

= 5 \cdot \int_{- 6}^{- 1} \frac{1}{u} + \frac{6}{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_{- 6}^{- 1} \frac{1}{u} d u + \int_{- 6}^{- 1} \frac{6}{u ^{2}} d u)

\int_{- 6}^{- 1} \frac{1}{u} d u = - ln (6)

\int_{- 6}^{- 1} \frac{6}{u ^{2}} d u = 5

= 5 (- ln (6) + 5)

x \to 1 lim (2 x)

\frac{d}{d x} (a x^{- 2})

y = \frac{( x + 3 )}{( x + 1 ) ^{2}}, at x = 0

\frac{\partial}{\partial x} (3 e^{x} - 5 x e^{y})

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

What is the integral from 0 to 5 of (5t)/((t-6)^2) ?
The integral from 0 to 5 of (5t)/((t-6)^2) is 5(-ln(6)+5)