What is the integral from 0 to 7 of (8t)/((t-8)^2) ?

The integral from 0 to 7 of (8t)/((t-8)^2) is 8(-3ln(2)+7)

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integral from 0 to 7 of (8t)/((t-8)^2)

\int_{0}^{7} \frac{8 t}{( t - 8 ) ^{2}} d t

8 (- 3 ln (2) + 7)

Decimal

39.36446 \dots

Solution steps

\int_{0}^{7} \frac{8 t}{( t - 8 ) ^{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}^{7} \frac{t}{( t - 8 ) ^{2}} d t

Apply u-substitution

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

Expand

\frac{u + 8}{u ^{2}} : \frac{1}{u} + \frac{8}{u ^{2}}

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

\int_{- 8}^{- 1} \frac{1}{u} d u = - 3 ln (2)

\int_{- 8}^{- 1} \frac{8}{u ^{2}} d u = 7

= 8 (- 3 ln (2) + 7)

\frac{d}{d t} (- e^{- t})

\frac{d}{d x} (x^{2} 7^{x})

\int_{- 1}^{1} (2 1 - x^{2})^{2} d x

\int (\frac{2}{3 x}) d x

x \to \infty lim (- 2 x^{3})

What is the integral from 0 to 7 of (8t)/((t-8)^2) ?
The integral from 0 to 7 of (8t)/((t-8)^2) is 8(-3ln(2)+7)