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integral of (2x-3)^2sqrt((2x-3)^2+4)

Lösung

\int (2 x - 3)^{2} (2 x - 3)^{2} + 4 d x

8 (\frac{1}{64} (2 x - 3) (4 x^{2} - 12 x + 13)^{\frac{3}{2}} - \frac{1}{32} (4 ln (\frac{2 x - 3 + 4 x ^{2} - 12 x + 13}{2}) + (2 x - 3) 4 x^{2} - 12 x + 13)) + C

Schritte zur Lösung

\int (2 x - 3)^{2} (2 x - 3)^{2} + 4 d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Trigonometrische Substitution anwenden

= \frac{1}{2} \cdot \int 16 sec^{3} (v) tan^{2} (v) d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot 16 \cdot \int sec^{3} (v) tan^{2} (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{1}{2} \cdot 16 \cdot \int sec^{3} (v) (- 1 + sec^{2} (v)) d v

Multipliziere aus

sec^{3} (v) (- 1 + sec^{2} (v)) : - sec^{3} (v) + sec^{5} (v)

= \frac{1}{2} \cdot 16 \cdot \int - sec^{3} (v) + sec^{5} (v) d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot 16 (- \int sec^{3} (v) d v + \int sec^{5} (v) d v)

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

\int sec^{5} (v) d v = \frac{sec ^{4} ( v ) sin ( v )}{4} + \frac{3}{4} (\frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

= \frac{1}{2} \cdot 16 (- (\frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣) + \frac{sec ^{4} ( v ) sin ( v )}{4} + \frac{3}{4} (\frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣))

Ersetze zurück

= \frac{1}{2} \cdot 16 (- (\frac{1}{2} sec (arctan (\frac{1}{2} (2 x - 3))) tan (arctan (\frac{1}{2} (2 x - 3))) + \frac{1}{2} ln tan (arctan (\frac{1}{2} (2 x - 3))) + sec (arctan (\frac{1}{2} (2 x - 3)))) + \frac{sec ^{4} ( arctan ( \frac{1}{2} ( 2 x - 3 ) ) ) sin ( arctan ( \frac{1}{2} ( 2 x - 3 ) ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (\frac{1}{2} (2 x - 3))) tan (arctan (\frac{1}{2} (2 x - 3))) + \frac{1}{2} ln tan (arctan (\frac{1}{2} (2 x - 3))) + sec (arctan (\frac{1}{2} (2 x - 3)))))

Vereinfache

\frac{1}{2} \cdot 16 (- (\frac{1}{2} sec (arctan (\frac{1}{2} (2 x - 3))) tan (arctan (\frac{1}{2} (2 x - 3))) + \frac{1}{2} ln tan (arctan (\frac{1}{2} (2 x - 3))) + sec (arctan (\frac{1}{2} (2 x - 3)))) + \frac{sec ^{4} ( arctan ( \frac{1}{2} ( 2 x - 3 ) ) ) sin ( arctan ( \frac{1}{2} ( 2 x - 3 ) ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (\frac{1}{2} (2 x - 3))) tan (arctan (\frac{1}{2} (2 x - 3))) + \frac{1}{2} ln tan (arctan (\frac{1}{2} (2 x - 3))) + sec (arctan (\frac{1}{2} (2 x - 3))))) : 8 (\frac{1}{64} (2 x - 3) (4 x^{2} - 12 x + 13)^{\frac{3}{2}} - \frac{1}{32} (4 ln (\frac{2 x - 3 + 4 x ^{2} - 12 x + 13}{2}) + (2 x - 3) 4 x^{2} - 12 x + 13))

= 8 (\frac{1}{64} (2 x - 3) (4 x^{2} - 12 x + 13)^{\frac{3}{2}} - \frac{1}{32} (4 ln (\frac{2 x - 3 + 4 x ^{2} - 12 x + 13}{2}) + (2 x - 3) 4 x^{2} - 12 x + 13))

Füge eine Konstante zur Lösung hinzu

= 8 (\frac{1}{64} (2 x - 3) (4 x^{2} - 12 x + 13)^{\frac{3}{2}} - \frac{1}{32} (4 ln (\frac{2 x - 3 + 4 x ^{2} - 12 x + 13}{2}) + (2 x - 3) 4 x^{2} - 12 x + 13)) + C

l a pl a ce t r an s f or m cos^{2} (17 t)

x \to \infty lim (\frac{x ^{l n (n)}}{x ^{l n (n) + 1}})

d er i v a t i v e ln (\frac{x}{y})

\int sin (2 x) sin (6 x) d x

\int \frac{x ^{2} - 9}{x ^{3}} d x