integral from 16 to 25 of (sqrt(x-16))/x

Lösung

\int_{16}^{25} \frac{x - 16}{x} d x

8 (- arctan (\frac{3}{4}) + \frac{3}{4})

Dezimale

0.85199 \dots

Schritte zur Lösung

\int_{16}^{25} \frac{x - 16}{x} d x

Wende U-Substitution an

= \int_{0}^{3} \frac{2 u ^{2}}{u ^{2} + 16} d u

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{3} \frac{u ^{2}}{u ^{2} + 16} d u

\frac{u ^{2}}{u ^{2} + 16} = - \frac{16}{u ^{2} + 16} + 1

= 2 \cdot \int_{0}^{3} - \frac{16}{u ^{2} + 16} + 1 d u

Wende integrale Substitution an

= 2 \cdot \int_{0}^{\frac{3}{4}} 4 (- \frac{1}{v ^{2} + 1} + 1) d v

Entferne die Konstante:

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

= 2 \cdot 4 \cdot \int_{0}^{\frac{3}{4}} - \frac{1}{v ^{2} + 1} + 1 d v

Wende die Summenregel an:

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

= 2 \cdot 4 (- \int_{0}^{\frac{3}{4}} \frac{1}{v ^{2} + 1} d v + \int_{0}^{\frac{3}{4}} 1 d v)

\int_{0}^{\frac{3}{4}} \frac{1}{v ^{2} + 1} d v = arctan (\frac{3}{4})

\int_{0}^{\frac{3}{4}} 1 d v = \frac{3}{4}

= 2 \cdot 4 (- arctan (\frac{3}{4}) + \frac{3}{4})

Vereinfache

= 8 (- arctan (\frac{3}{4}) + \frac{3}{4})

\int cos (x) (8 + 5 sin^{2} (x)) d x

\int \frac{1}{x ^{2} + 6 x + 8} d x

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

\frac{\partial}{\partial x} (sin (π (6 x - 2 y)))

\int_{1}^{\infty} \frac{1}{x ^{2} + 2} d x