integral from 0 to 3 of (5t)/((t-4)^2)

Lösung

\int_{0}^{3} \frac{5 t}{( t - 4 ) ^{2}} d t

5 (- 2 ln (2) + 3)

Dezimale

8.06852 \dots

Schritte zur Lösung

\int_{0}^{3} \frac{5 t}{( t - 4 ) ^{2}} d t

Entferne die Konstante:

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

= 5 \cdot \int_{0}^{3} \frac{t}{( t - 4 ) ^{2}} d t

Wende U-Substitution an

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

Multipliziere aus

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

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

= 5 (\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

= 5 (- 2 ln (2) + 3)

\frac{d y}{d x} = \frac{cos ( 15 x )}{e ^{15 y}}

9 t (\frac{d y}{d t}) + y = t^{2}

\frac{\partial}{\partial y} (\frac{e ^{2 y}}{2 x})

\int_{1}^{e} 4 x^{2} ln (x) d x

\frac{d}{d x} (\frac{1}{1 + x})