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integral of (sqrt(x+16))/x

Lösung

\int \frac{x + 16}{x} d x

- 8 (\frac{1}{2} ln \frac{x + 16}{4} + 1 - \frac{1}{2} ln \frac{x + 16}{4} - 1 - \frac{1}{4} x + 16) + C

Schritte zur Lösung

\int \frac{x + 16}{x} d x

Wende U-Substitution an

= \int \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 \frac{u ^{2}}{u ^{2} - 16} d u

Faktorisiere

u^{2} - 16 : - (- u^{2} + 16)

= 2 \cdot \int \frac{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 (- \int \frac{u ^{2}}{- u ^{2} + 16} d u)

Wende integrale Substitution an

= 2 (- \int \frac{4 v ^{2}}{- v ^{2} + 1} d v)

Entferne die Konstante:

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

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

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

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

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

\int 1 d v = v

= 2 (- 4 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v))

Ersetze zurück

= 2 - 4 \frac{ln \frac{x + 16}{4} + 1}{2} - \frac{ln \frac{x + 16}{4} - 1}{2} - \frac{x + 16}{4}

Vereinfache

2 - 4 \frac{ln \frac{x + 16}{4} + 1}{2} - \frac{ln \frac{x + 16}{4} - 1}{2} - \frac{x + 16}{4} : - 8 (\frac{1}{2} ln \frac{x + 16}{4} + 1 - \frac{1}{2} ln \frac{x + 16}{4} - 1 - \frac{1}{4} x + 16)

= - 8 (\frac{1}{2} ln \frac{x + 16}{4} + 1 - \frac{1}{2} ln \frac{x + 16}{4} - 1 - \frac{1}{4} x + 16)

Füge eine Konstante zur Lösung hinzu

= - 8 (\frac{1}{2} ln \frac{x + 16}{4} + 1 - \frac{1}{2} ln \frac{x + 16}{4} - 1 - \frac{1}{4} x + 16) + C

x \to - 3 lim (- (3 x))

d er i v a t i v e e^{(3 h^{2} + 6)}

t \to 0 lim (\frac{( e ^{t} - 1 )}{t})

\frac{d}{d x} ((x + y)^{2} + (x + 1)^{2})

in v erse l a pl a ce \frac{- 20}{s ( s + 10 )}