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

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

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

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

8 (- ln (7) + 6)

Decimal Notation

32.43271 \dots

Solution steps

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

Apply u-substitution

: \int_{- 7}^{- 1} \frac{u + 7}{u ^{2}} d u

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

Expand

\frac{u + 7}{u ^{2}} : \frac{1}{u} + \frac{7}{u ^{2}}

= 8 \cdot \int_{- 7}^{- 1} \frac{1}{u} + \frac{7}{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_{- 7}^{- 1} \frac{1}{u} d u + \int_{- 7}^{- 1} \frac{7}{u ^{2}} d u)

\int_{- 7}^{- 1} \frac{1}{u} d u = - ln (7)

\int_{- 7}^{- 1} \frac{7}{u ^{2}} d u = 6

= 8 (- ln (7) + 6)

\frac{\partial}{\partial x} (\frac{z ( x + y )}{x + y + z})

x^{^{'}} = - 2 x

x \to - 10 lim (- (\frac{1}{x + 10}))

\int_{0}^{x} x^{2} + x^{4} d x

\frac{d}{d x} (a x^{2} e^{3 x})

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