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

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

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

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

7 (- ln (7) + 6)

Decimal Notation

28.37862 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

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

= 7 \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

= 7 (\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

= 7 (- ln (7) + 6)

\frac{e ^{- 4 s}}{s ^{2}}

98.284 e^{- 0.28 x}

\int_{0}^{1} 1 0^{x} d x

x \to 2 lim (4 tan (3 x - 3))

\frac{d y}{d x}, x cos (5 y) + y^{5} e^{6 x} = e^{2 x}

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