What is the integral from 0 to 6 of (5t)/((t-7)^2) ?

The integral from 0 to 6 of (5t)/((t-7)^2) is 5(-ln(7)+6)

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integral from 0 to 6 of (5t)/((t-7)^2)

\int_{0}^{6} \frac{5 t}{( t - 7 ) ^{2}} d t

5 (- ln (7) + 6)

Decimal Notation

20.27044 \dots

Solution steps

\int_{0}^{6} \frac{5 t}{( t - 7 ) ^{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}^{6} \frac{t}{( t - 7 ) ^{2}} d t

Apply u-substitution

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

= 5 \cdot \int_{- 7}^{- 1} \frac{u + 7}{u ^{2}} d u

Expand

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

= 5 \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

= 5 (\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

= 5 (- ln (7) + 6)

\int \frac{1}{( 4 - x ^{2} ) ^{3}} d x

\int ln (x + 1) x d x

\frac{d}{d x} (e x)

\frac{\partial}{\partial v} (\frac{u}{u + v})

y^{^{'}} = e^{6 x} - 4 x

What is the integral from 0 to 6 of (5t)/((t-7)^2) ?
The integral from 0 to 6 of (5t)/((t-7)^2) is 5(-ln(7)+6)