What is the integral from 0 to 3 of (8t)/((t-4)^2) ?

The integral from 0 to 3 of (8t)/((t-4)^2) is 8(-2ln(2)+3)

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integral from 0 to 3 of (8t)/((t-4)^2)

\int_{0}^{3} \frac{8 t}{( t - 4 ) ^{2}} d t

8 (- 2 ln (2) + 3)

Decimal

12.90964 \dots

Solution steps

\int_{0}^{3} \frac{8 t}{( t - 4 ) ^{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}^{3} \frac{t}{( t - 4 ) ^{2}} d t

Apply u-substitution

= 8 \cdot \int_{- 4}^{- 1} \frac{u + 4}{u ^{2}} d u

Expand

\frac{u + 4}{u ^{2}} : \frac{1}{u} + \frac{4}{u ^{2}}

= 8 \cdot \int_{- 4}^{- 1} \frac{1}{u} + \frac{4}{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_{- 4}^{- 1} \frac{1}{u} d u + \int_{- 4}^{- 1} \frac{4}{u ^{2}} d u)

\int_{- 4}^{- 1} \frac{1}{u} d u = - 2 ln (2)

\int_{- 4}^{- 1} \frac{4}{u ^{2}} d u = 3

= 8 (- 2 ln (2) + 3)

What is the integral from 0 to 3 of (8t)/((t-4)^2) ?
The integral from 0 to 3 of (8t)/((t-4)^2) is 8(-2ln(2)+3)