integral of sqrt(4r^2+1)

Lösung

\int 4 r^{2} + 1 d r

\frac{1}{4} (2 r 4 r^{2} + 1 + ln 2 r + 1 + 4 r^{2}) + C

Schritte zur Lösung

\int 4 r^{2} + 1 d r

Trigonometrische Substitution anwenden

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

Wende integrale Reduktion an

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

Setze in

u = arctan (2 r)

ein

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

Vereinfache

\frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 r ) ) sin ( arctan ( 2 r ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 r)) + sec (arctan (2 r)) ∣) : \frac{1}{4} (2 r 4 r^{2} + 1 + ln 2 r + 1 + 4 r^{2})

= \frac{1}{4} (2 r 4 r^{2} + 1 + ln 2 r + 1 + 4 r^{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (2 r 4 r^{2} + 1 + ln 2 r + 1 + 4 r^{2}) + C

\int_{- \frac{1}{2}}^{0} (\frac{1}{2} + 2 y + 3 z) d y

\frac{\partial}{\partial x} (e^{x s i n (y)})

t a y l or 1 + 2 x

x \to 2 lim (\frac{3 x + 4}{8 x - 1})

\int \frac{1}{u 3 - u ^{2}} d u