integral of sqrt(7+x^2)

Lösung

\int 7 + x^{2} d x

7 (\frac{1}{14} x 7 + x^{2} + \frac{1}{2} ln \frac{1}{7} x + \frac{1}{7} (7 + x^{2})) + C

Schritte zur Lösung

\int 7 + x^{2} d x

Trigonometrische Substitution anwenden

= \int 7 sec^{3} (u) d u

Entferne die Konstante:

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

= 7 \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= 7 (\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) ∣

= 7 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arctan (\frac{1}{7} x)

ein

= 7 \frac{sec ^{2} ( arctan ( \frac{1}{7} x ) ) sin ( arctan ( \frac{1}{7} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{7} x)) + sec (arctan (\frac{1}{7} x))

Vereinfache

7 \frac{sec ^{2} ( arctan ( \frac{1}{7} x ) ) sin ( arctan ( \frac{1}{7} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{7} x)) + sec (arctan (\frac{1}{7} x)) : 7 (\frac{1}{14} x 7 + x^{2} + \frac{1}{2} ln \frac{1}{7} x + \frac{1}{7} (7 + x^{2}))

= 7 (\frac{1}{14} x 7 + x^{2} + \frac{1}{2} ln \frac{1}{7} x + \frac{1}{7} (7 + x^{2}))

Füge eine Konstante zur Lösung hinzu

= 7 (\frac{1}{14} x 7 + x^{2} + \frac{1}{2} ln \frac{1}{7} x + \frac{1}{7} (7 + x^{2})) + C

x^{^{''}} - 0.1 x^{^{'}} + x = 0

\int (\frac{x}{2})^{2} d x

t an g e n t f (x) = x^{4} - 8 x^{3} + 4, at x = - 1

(x^{2} + \frac{2 y}{x}) d x = (3 - ln (x^{2})) d y

\frac{d}{d x} ((2 x^{5} - 5 x^{3} + 7 x)^{4})