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

Lösung

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

\frac{1}{4} (2 (x - 2) 4 x^{2} - 16 x + 17 + ln 2 (x - 2) + 4 x^{2} - 16 x + 17) + C

Schritte zur Lösung

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

Multipliziere aus

1 + (2 x - 4)^{2} : 4 x^{2} - 16 x + 17

= \int 4 x^{2} - 16 x + 17 d x

Vervollständige das Quadrat

4 x^{2} - 16 x + 17 : 4 (x - 2)^{2} + 1

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

Wende U-Substitution an

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

Trigonometrische Substitution anwenden

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

Wende integrale Reduktion an

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

Ersetze zurück

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

Vereinfache

\frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 ( x - 2 ) ) ) sin ( arctan ( 2 ( x - 2 ) ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 (x - 2))) + sec (arctan (2 (x - 2))) ∣) : \frac{1}{4} (2 (x - 2) 4 x^{2} - 16 x + 17 + ln 2 (x - 2) + 4 x^{2} - 16 x + 17)

= \frac{1}{4} (2 (x - 2) 4 x^{2} - 16 x + 17 + ln 2 (x - 2) + 4 x^{2} - 16 x + 17)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (2 (x - 2) 4 x^{2} - 16 x + 17 + ln 2 (x - 2) + 4 x^{2} - 16 x + 17) + C

\int 3 x e^{x^{3}} d x

\int x (x^{2} - 16)^{2} d x

\int \frac{3 x + 2}{x ^{2} + 4} d x

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

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