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Beliebt Rechnen >

integral of sqrt(1+u^2)

Lösung

\int 1 + u^{2} d u

Lösung

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

Schritte zur Lösung

\int 1 + u^{2} d u

Trigonometrische Substitution anwenden

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

Wende integrale Reduktion an

= \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) ∣

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

Setze in

v = arctan (u)

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

Graph

Beliebte Beispiele

integral from 0 to 4 of x^3sqrt(x^2+16)

\int_{0}^{4} x^{3} x^{2} + 16 d x

derivative of 3x^2+2x+1

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

t^{2} y^{^{'}} + t y = 5

laplacetransform xsin(3x)

l a pl a ce t r an s f or m x sin (3 x)

derivative of 1/(e^x+1)

\frac{d}{d x} (\frac{1}{e ^{x} + 1})