integral (1+x^2)^{0.5}

Lösung

integral

(1 + x^{2})^{0.5}

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

Schritte zur Lösung

\int (1 + x^{2})^{0.5} d x

Trigonometrische Substitution anwenden

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

Wende integrale Reduktion an

= \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) ∣

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

Setze in

u = arctan (x)

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l \frac{tan ^{2} ( x )}{cos ( x )}

in t e g r a l \frac{( 4 - x )}{x}

in t e g r a l \frac{( e ^{- t} - 4 )}{e ^{- t}}

in t e g r a l e^{6} x cos (e^{3} x)

in t e g r a l 2 x + 5