integral of sqrt(3+x^2)

Lösung

\int 3 + x^{2} d x

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

Schritte zur Lösung

\int 3 + x^{2} d x

Trigonometrische Substitution anwenden

= \int 3 sec^{3} (u) d u

Entferne die Konstante:

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

= 3 \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

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

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

= 3 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arctan (\frac{1}{3} x)

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int (3 x^{2} - 9 x + 2) d x

\int \frac{1 + cos ( x )}{1 - cos ( x )} d x

\frac{d}{d x} (arcsin (5 x))

d er i v a t i v e \frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{1}{3}} + 9

x \to - 2 + lim (\frac{1 - x}{x ^{3} + 8})