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

Lösung

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

2 (- ln (5) + 4)

Dezimale

4.78112 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende U-Substitution an

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

Multipliziere aus

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

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\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

= 2 (- ln (5) + 4)

x \to 0 lim (ln (1 + x))

\frac{\partial}{\partial z} (8 x^{\frac{y}{z}})

l a pl a ce t r an s f or m 3 (t - 2)^{2} + 13 (t - 2) + 17

n = 0 \sum \infty n (1 - p)^{n - 1}

\int_{1}^{3} \frac{1}{x} d x