What is the integral from 0 to 4 of (4t)/((t-5)^2) ?

The integral from 0 to 4 of (4t)/((t-5)^2) is 4(-ln(5)+4)

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integral from 0 to 4 of (4t)/((t-5)^2)

\int_{0}^{4} \frac{4 t}{( t - 5 ) ^{2}} d t

4 (- ln (5) + 4)

Solution steps

\int_{0}^{4} \frac{4 t}{( t - 5 ) ^{2}} d t

Take the constant out:

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

= 4 \cdot \int_{0}^{4} \frac{t}{( t - 5 ) ^{2}} d t

Apply u-substitution

: \int_{- 5}^{- 1} \frac{u + 5}{u ^{2}} d u

= 4 \cdot \int_{- 5}^{- 1} \frac{u + 5}{u ^{2}} d u

Expand

\frac{u + 5}{u ^{2}} : \frac{1}{u} + \frac{5}{u ^{2}}

= 4 \cdot \int_{- 5}^{- 1} \frac{1}{u} + \frac{5}{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

= 4 (\int_{- 5}^{- 1} \frac{1}{u} d u + \int_{- 5}^{- 1} \frac{5}{u ^{2}} d u)

\int_{- 5}^{- 1} \frac{1}{u} d u = - ln (5)

\int_{- 5}^{- 1} \frac{5}{u ^{2}} d u = 4

= 4 (- ln (5) + 4)

y^{^{'}} + 4 y = 8

t (5 t^{3} + 8 t^{2} + 10)

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

\int_{0}^{0.1} x arctan (5 x) d x

\int 8 x^{\frac{3}{5}} + 5 x^{- \frac{4}{5}} d x

What is the integral from 0 to 4 of (4t)/((t-5)^2) ?
The integral from 0 to 4 of (4t)/((t-5)^2) is 4(-ln(5)+4)