What is the integral from 0 to 7 of (5t)/((t-8)^2) ?

The integral from 0 to 7 of (5t)/((t-8)^2) is 5(-3ln(2)+7)

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integral from 0 to 7 of (5t)/((t-8)^2)

\int_{0}^{7} \frac{5 t}{( t - 8 ) ^{2}} d t

5 (- 3 ln (2) + 7)

Decimal Notation

24.60279 \dots

Solution steps

\int_{0}^{7} \frac{5 t}{( t - 8 ) ^{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}^{7} \frac{t}{( t - 8 ) ^{2}} d t

Apply u-substitution

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

= 5 \cdot \int_{- 8}^{- 1} \frac{u + 8}{u ^{2}} d u

Expand

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

= 5 \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

= 5 (\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

= 5 (- 3 ln (2) + 7)

\frac{\partial}{\partial x} (e^{(- 0.1 x y)})

\frac{\partial}{\partial x} (e^{x} cos (y) + yz)

\frac{d}{d x} (\frac{x ( x + 10 )}{( x + 5 ) ^{2}})

x \to 1 lim ((2 x - 3) \cdot (x + 1))

y^{^{''}} + 1.5 y^{^{'}} - y = 12 x^{2} + 6 x^{3} - x^{4}

What is the integral from 0 to 7 of (5t)/((t-8)^2) ?
The integral from 0 to 7 of (5t)/((t-8)^2) is 5(-3ln(2)+7)