integral of 1/(xsqrt(16x^2+1))

Lösung

\int \frac{1}{x 16 x ^{2} + 1} d x

- \frac{1}{2} ln 16 x^{2} + 1 + 1 + \frac{1}{2} ln 16 x^{2} + 1 - 1 + C

Schritte zur Lösung

\int \frac{1}{x 16 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 = 16 x^{2} + 1

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

im pl i c i t \frac{d y}{d x}, ln (y) = x

\int \frac{y}{1} d y

\frac{d}{d t} (1 - ∣ t ∣)

\int - sin (x) - e^{- x} + 3 x^{2} d x

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x^{2} + e^{- x}