What is the integral from 0 to 5 of (8t)/((t-6)^2) ?

The integral from 0 to 5 of (8t)/((t-6)^2) is 8(-ln(6)+5)

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integral from 0 to 5 of (8t)/((t-6)^2)

\int_{0}^{5} \frac{8 t}{( t - 6 ) ^{2}} d t

8 (- ln (6) + 5)

Decimal Notation

25.66592 \dots

Solution steps

\int_{0}^{5} \frac{8 t}{( t - 6 ) ^{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}^{5} \frac{t}{( t - 6 ) ^{2}} d t

Apply u-substitution

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

= 8 \cdot \int_{- 6}^{- 1} \frac{u + 6}{u ^{2}} d u

Expand

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

= 8 \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

= 8 (\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

= 8 (- ln (6) + 5)

\int x^{\frac{2}{5}} d x

x \to - 2 lim (e^{2 x - x^{2}})

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

f (x) = \frac{x - 1}{2 - x}, at x = 4

\int (x^{3} - 3 x + x^{- \frac{5}{4}}) d x

What is the integral from 0 to 5 of (8t)/((t-6)^2) ?
The integral from 0 to 5 of (8t)/((t-6)^2) is 8(-ln(6)+5)