integral of x/(x^2sqrt(x^2+1))

Lösung

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

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

Schritte zur Lösung

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

Streiche die gemeinsamen Faktoren:

x

= \int \frac{1}{x x ^{2} + 1} d x

Wende U-Substitution an

= \int \frac{1}{u ^{2} - 1} d u

Faktorisiere

u^{2} - 1 : - (- u^{2} + 1)

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

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

Setze in

u = x^{2} + 1

ein

= - (\frac{ln x ^{2} + 1 + 1}{2} - \frac{ln x ^{2} + 1 - 1}{2})

Vereinfache

- (\frac{ln x ^{2} + 1 + 1}{2} - \frac{ln x ^{2} + 1 - 1}{2}) : - \frac{1}{2} ln x^{2} + 1 + 1 + \frac{1}{2} ln x^{2} + 1 - 1

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

Füge eine Konstante zur Lösung hinzu

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

\int ((8 - x) (8 + x)) d x

\int x^{3} ln (x) + cos (3 x + 2) d x

\int \frac{4 x ^{2}}{7 x ^{3} + 6} d x

\int \frac{4}{8 + 5 x} d x

\int 3 x e^{- 5 x} d x