integral from 0 to 7 of (7t)/((t-8)^2)

Lösung

\int_{0}^{7} \frac{7 t}{( t - 8 ) ^{2}} d t

7 (- 3 ln (2) + 7)

Dezimale

34.44390 \dots

Schritte zur Lösung

\int_{0}^{7} \frac{7 t}{( t - 8 ) ^{2}} d t

Entferne die Konstante:

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

= 7 \cdot \int_{0}^{7} \frac{t}{( t - 8 ) ^{2}} d t

Wende U-Substitution an

= 7 \cdot \int_{- 8}^{- 1} \frac{u + 8}{u ^{2}} d u

Multipliziere aus

\frac{u + 8}{u ^{2}} : \frac{1}{u} + \frac{8}{u ^{2}}

= 7 \cdot \int_{- 8}^{- 1} \frac{1}{u} + \frac{8}{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

= 7 (\int_{- 8}^{- 1} \frac{1}{u} d u + \int_{- 8}^{- 1} \frac{8}{u ^{2}} d u)

\int_{- 8}^{- 1} \frac{1}{u} d u = - 3 ln (2)

\int_{- 8}^{- 1} \frac{8}{u ^{2}} d u = 7

= 7 (- 3 ln (2) + 7)

s im pl i f y - 3

f (x) = arccos (e^{x})

\int 2 e^{x} - 6 x d x

6 t \frac{d y}{d t} + y = 0

\frac{d}{d x} (\frac{1}{3} sin^{3} (x))