integral of sqrt(1+64x^2)

Lösung

\int 1 + 64 x^{2} d x

\frac{1}{16} (8 x 64 x^{2} + 1 + ln 8 x + 1 + 64 x^{2}) + C

Schritte zur Lösung

\int 1 + 64 x^{2} d x

Trigonometrische Substitution anwenden

= \int \frac{1}{8} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{1}{8} (\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}{8} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arctan (8 x)

ein

= \frac{1}{8} (\frac{sec ^{2} ( arctan ( 8 x ) ) sin ( arctan ( 8 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (8 x)) + sec (arctan (8 x)) ∣)

Vereinfache

\frac{1}{8} (\frac{sec ^{2} ( arctan ( 8 x ) ) sin ( arctan ( 8 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (8 x)) + sec (arctan (8 x)) ∣) : \frac{1}{16} (8 x 64 x^{2} + 1 + ln 8 x + 1 + 64 x^{2})

= \frac{1}{16} (8 x 64 x^{2} + 1 + ln 8 x + 1 + 64 x^{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{16} (8 x 64 x^{2} + 1 + ln 8 x + 1 + 64 x^{2}) + C

\int e^{2} cos (x) d x

in v erse l a pl a ce \frac{4}{s ^{4} + 4}

\int (2 x + 3)^{10} d x

y^{^{''}} + 9 y^{^{'}} + 20 y = 0

a re a 4 + x, 4 + \frac{1}{5} x