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

Lösung

integral

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

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

Schritte zur Lösung

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

Vereinfache

x^{2} : xx

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

xx = x^{2}

= \int (x^{2} + 1)^{\frac{1}{2}} 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{1}{( x + a )}

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

in t e g r a l \frac{2}{( 1 + x ^{2} )}

in t e g r a l \frac{ln ( t )}{( t + 1 ) ^{2}}

in t e g r a l (u^{2}) (1 + u^{2})^{0.5}