integral of xsec(3x^2+1)

Lösung

\int x sec (3 x^{2} + 1) d x

\frac{1}{6} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1) + C

Schritte zur Lösung

\int x sec (3 x^{2} + 1) d x

Wende U-Substitution an

= \int \frac{sec ( 3 u + 1 )}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int sec (3 u + 1) d u

Wende U-Substitution an

= \frac{1}{2} \cdot \int sec (v) \frac{1}{3} d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot \frac{1}{3} \cdot \int sec (v) d v

Nutze das gemeinsame Integral :

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

= \frac{1}{2} \cdot \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= \frac{1}{2} \cdot \frac{1}{3} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1)

Vereinfache

\frac{1}{2} \cdot \frac{1}{3} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1) : \frac{1}{6} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1)

= \frac{1}{6} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1) + C

6 \frac{d y}{d x} - 2 y = x y^{4}, y (0) = - 2

(\frac{2 y}{x} + 2 x) + (2 ln (x) - 3) (\frac{d y}{d x}) = 0

d er i v a t i v e \frac{1}{4 ^{x}}

\int_{\frac{1}{2}}^{1} \frac{4}{9} x (5 - x^{2}) d x

d er i v a t i v e \frac{t - t}{t ^{\frac{1}{7}}}