integral of (sqrt(36-x^2))/x

Lösung

\int \frac{36 - x ^{2}}{x} d x

- 3 ln \frac{1}{6} - x^{2} + 36 + 1 + 3 ln \frac{1}{6} - x^{2} + 36 - 1 + 36 - x^{2} + C

Schritte zur Lösung

\int \frac{36 - x ^{2}}{x} d x

Wende U-Substitution an

= \int - \frac{u ^{2}}{- u ^{2} + 36} d u

Entferne die Konstante:

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

= - \int \frac{u ^{2}}{- u ^{2} + 36} d u

Wende integrale Substitution an

= - \int \frac{6 v ^{2}}{- v ^{2} + 1} d v

Entferne die Konstante:

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

= - 6 \cdot \int \frac{v ^{2}}{- v ^{2} + 1} d v

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

= - 6 \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

= - 6 (\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

= - 6 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v)

Ersetze zurück

= - 6 \frac{ln \frac{36 - x ^{2}}{6} + 1}{2} - \frac{ln \frac{36 - x ^{2}}{6} - 1}{2} - \frac{36 - x ^{2}}{6}

Vereinfache

- 6 \frac{ln \frac{36 - x ^{2}}{6} + 1}{2} - \frac{ln \frac{36 - x ^{2}}{6} - 1}{2} - \frac{36 - x ^{2}}{6} : - 3 ln \frac{1}{6} - x^{2} + 36 + 1 + 3 ln \frac{1}{6} - x^{2} + 36 - 1 + 36 - x^{2}

= - 3 ln \frac{1}{6} - x^{2} + 36 + 1 + 3 ln \frac{1}{6} - x^{2} + 36 - 1 + 36 - x^{2}

Füge eine Konstante zur Lösung hinzu

= - 3 ln \frac{1}{6} - x^{2} + 36 + 1 + 3 ln \frac{1}{6} - x^{2} + 36 - 1 + 36 - x^{2} + C

y^{^{''}} = 36 x^{7} + 21 x^{5} - 3 x + 48 x^{- 5}

ma c l a u r in sin (x), 1

\frac{d}{d x} (3 sin (2 x) + cos (2 x))

t an g e n t f (x) = \frac{8}{x ^{2} + x + 2}, at x = 2

h \to + 0 lim (2 x + h)