What is the integral from 0 to 7 of (6t)/((t-8)^2) ?

The integral from 0 to 7 of (6t)/((t-8)^2) is 6(-3ln(2)+7)

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integral from 0 to 7 of (6t)/((t-8)^2)

\int_{0}^{7} \frac{6 t}{( t - 8 ) ^{2}} d t

6 (- 3 ln (2) + 7)

Decimal Notation

29.52335 \dots

Solution steps

\int_{0}^{7} \frac{6 t}{( t - 8 ) ^{2}} d t

Take the constant out:

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

= 6 \cdot \int_{0}^{7} \frac{t}{( t - 8 ) ^{2}} d t

Apply u-substitution

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

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

Expand

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

= 6 \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

= 6 (\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

= 6 (- 3 ln (2) + 7)

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

n = 0 \sum \infty n (\frac{3}{4})^{n}

\frac{\partial}{\partial y} (\frac{- y}{( x ^{2} + y ^{2} )})

(ln (x + 1))^{2} - e^{x^{2}}

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

What is the integral from 0 to 7 of (6t)/((t-8)^2) ?
The integral from 0 to 7 of (6t)/((t-8)^2) is 6(-3ln(2)+7)