integral of (2sqrt(ln^2(x)+16))/x

Lösung

\int \frac{2 ln ^{2} ( x ) + 16}{x} d x

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

Schritte zur Lösung

\int \frac{2 ln ^{2} ( x ) + 16}{x} d x

Entferne die Konstante:

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

= 2 \cdot \int \frac{ln ^{2} ( x ) + 16}{x} d x

Wende U-Substitution an

= 2 \cdot \int u^{2} + 16 d u

Trigonometrische Substitution anwenden

= 2 \cdot \int 16 sec^{3} (v) d v

Entferne die Konstante:

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

= 2 \cdot 16 \cdot \int sec^{3} (v) d v

Wende integrale Reduktion an

= 2 \cdot 16 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 2 \cdot 16 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Ersetze zurück

= 2 \cdot 16 (\frac{sec ^{2} ( arctan ( \frac{1}{4} ln ( x ) ) ) sin ( arctan ( \frac{1}{4} ln ( x ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{4} ln (x))) + sec (arctan (\frac{1}{4} ln (x))))

Vereinfache

2 \cdot 16 (\frac{sec ^{2} ( arctan ( \frac{1}{4} ln ( x ) ) ) sin ( arctan ( \frac{1}{4} ln ( x ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{4} ln (x))) + sec (arctan (\frac{1}{4} ln (x)))) : ln (x) 16 + ln^{2} (x) + 16 ln (\frac{1}{4} ln (x) + 16 + ln^{2} (x))

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

Füge eine Konstante zur Lösung hinzu

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

\int 4 t + 2 d t

d er i v a t i v e θ

in v erse l a pl a ce e^{s}

d er i v a t i v e 500 q - \frac{q ^{2}}{ln ( q + 300 )}

(x + 2 y) d x + x d y = 0