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

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

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

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

7 (- ln (6) + 5)

Decimal

22.45768 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{- 6}^{- 1} \frac{u + 6}{u ^{2}} d u

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

Expand

\frac{u + 6}{u ^{2}} : \frac{1}{u} + \frac{6}{u ^{2}}

= 7 \cdot \int_{- 6}^{- 1} \frac{1}{u} + \frac{6}{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

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

\int_{- 6}^{- 1} \frac{1}{u} d u = - ln (6)

\int_{- 6}^{- 1} \frac{6}{u ^{2}} d u = 5

= 7 (- ln (6) + 5)

\frac{d y}{d x} = \frac{4 x}{y}

f (x) = \frac{1}{x ^{3}}, (- 3, - \frac{1}{27})

\int \frac{t ^{2}}{( t ^{3} - 2 ) ^{5}} d t

\frac{d}{d x} (x^{0.85})

n = 1 \sum \infty (1 - \frac{1}{3 n})^{n}

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