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integral of sqrt(1+(x/2-4)^2)

Lösung

\int 1 + (\frac{x}{2} - 4)^{2} d x

ln (\frac{1}{2} x - 8 + x^{2} - 16 x + 68) + \frac{1}{4} (x - 8) x^{2} - 16 x + 68 + C

Schritte zur Lösung

\int 1 + (\frac{x}{2} - 4)^{2} d x

Vereinfache

1 + (\frac{x}{2} - 4)^{2} : \frac{1}{2} x^{2} - 16 x + 68

= \int \frac{1}{2} x^{2} - 16 x + 68 d x

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int x^{2} - 16 x + 68 d x

Vervollständige das Quadrat

x^{2} - 16 x + 68 : (x - 8)^{2} + 4

= \frac{1}{2} \cdot \int (x - 8)^{2} + 4 d x

Wende U-Substitution an

= \frac{1}{2} \cdot \int u^{2} + 4 d u

Trigonometrische Substitution anwenden

= \frac{1}{2} \cdot \int 4 sec^{3} (v) d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot 4 \cdot \int sec^{3} (v) d v

Wende integrale Reduktion an

= \frac{1}{2} \cdot 4 (\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{1}{2} \cdot 4 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Ersetze zurück

= \frac{1}{2} \cdot 4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} ( x - 8 ) ) ) sin ( arctan ( \frac{1}{2} ( x - 8 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} (x - 8))) + sec (arctan (\frac{1}{2} (x - 8))))

Vereinfache

\frac{1}{2} \cdot 4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} ( x - 8 ) ) ) sin ( arctan ( \frac{1}{2} ( x - 8 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} (x - 8))) + sec (arctan (\frac{1}{2} (x - 8)))) : ln (\frac{1}{2} x - 8 + x^{2} - 16 x + 68) + \frac{1}{4} (x - 8) x^{2} - 16 x + 68

= ln (\frac{1}{2} x - 8 + x^{2} - 16 x + 68) + \frac{1}{4} (x - 8) x^{2} - 16 x + 68

Füge eine Konstante zur Lösung hinzu

= ln (\frac{1}{2} x - 8 + x^{2} - 16 x + 68) + \frac{1}{4} (x - 8) x^{2} - 16 x + 68 + C

\int 3 + 2 cos (x) d x

θ \to 0 lim (\frac{sin ( 3 θ )}{θ})

x \to 3 lim (\frac{x ^{2}}{x ^{2} - 9})

a re a 4 x + y^{2} = 12, x = y

\int \frac{x ^{4} + x ^{2} - 1}{x - 1} d x