English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^2)/((15+6x-9x^2)^{3/2)}

Lösung

\int \frac{x ^{2}}{( 15 + 6 x - 9 x ^{2} ) ^{\frac{3}{2}}} d x

\frac{51 x - 16 arcsin ( \frac{3 x}{4} - \frac{1}{4} ) ( - 9 x ^{2} + 6 x + 15 ) ^{\frac{1}{2}} + 15}{432 ( - 9 x ^{2} + 6 x + 15 ) ^{\frac{1}{2}}} + C

Schritte zur Lösung

\int \frac{x ^{2}}{( 15 + 6 x - 9 x ^{2} ) ^{\frac{3}{2}}} d x

Vervollständige das Quadrat

15 + 6 x - 9 x^{2} : - 9 (x - \frac{1}{3})^{2} + 16

= \int \frac{x ^{2}}{( - 9 ( x - \frac{1}{3} ) ^{2} + 16 ) ^{\frac{3}{2}}} d x

Wende U-Substitution an

= \int \frac{( 3 u + 1 ) ^{2}}{9 ( - 9 u ^{2} + 16 ) ^{\frac{3}{2}}} d u

Entferne die Konstante:

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

= \frac{1}{9} \cdot \int \frac{( 3 u + 1 ) ^{2}}{( - 9 u ^{2} + 16 ) ^{\frac{3}{2}}} d u

Trigonometrische Substitution anwenden

= \frac{1}{9} \cdot \int \frac{( 4 sin ( v ) + 1 ) ^{2}}{48 cos ^{2} ( v )} d v

Entferne die Konstante:

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

= \frac{1}{9} \cdot \frac{1}{48} \cdot \int \frac{( 4 sin ( v ) + 1 ) ^{2}}{cos ^{2} ( v )} d v

Multipliziere aus

\frac{( 4 sin ( v ) + 1 ) ^{2}}{cos ^{2} ( v )} : \frac{16 sin ^{2} ( v )}{cos ^{2} ( v )} + \frac{8 sin ( v )}{cos ^{2} ( v )} + \frac{1}{cos ^{2} ( v )}

= \frac{1}{9} \cdot \frac{1}{48} \cdot \int \frac{16 sin ^{2} ( v )}{cos ^{2} ( v )} + \frac{8 sin ( v )}{cos ^{2} ( v )} + \frac{1}{cos ^{2} ( 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}{9} \cdot \frac{1}{48} (\int \frac{16 sin ^{2} ( v )}{cos ^{2} ( v )} d v + \int \frac{8 sin ( v )}{cos ^{2} ( v )} d v + \int \frac{1}{cos ^{2} ( v )} d v)

\int \frac{16 sin ^{2} ( v )}{cos ^{2} ( v )} d v = 16 (tan (v) - v)

\int \frac{8 sin ( v )}{cos ^{2} ( v )} d v = 8 sec (v)

\int \frac{1}{cos ^{2} ( v )} d v = tan (v)

= \frac{1}{9} \cdot \frac{1}{48} (16 (tan (v) - v) + 8 sec (v) + tan (v))

Ersetze zurück

= \frac{1}{9} \cdot \frac{1}{48} (16 (tan (arcsin (\frac{3}{4} (x - \frac{1}{3}))) - arcsin (\frac{3}{4} (x - \frac{1}{3}))) + 8 sec (arcsin (\frac{3}{4} (x - \frac{1}{3}))) + tan (arcsin (\frac{3}{4} (x - \frac{1}{3}))))

Vereinfache

\frac{1}{9} \cdot \frac{1}{48} (16 (tan (arcsin (\frac{3}{4} (x - \frac{1}{3}))) - arcsin (\frac{3}{4} (x - \frac{1}{3}))) + 8 sec (arcsin (\frac{3}{4} (x - \frac{1}{3}))) + tan (arcsin (\frac{3}{4} (x - \frac{1}{3})))) : \frac{51 x - 16 arcsin ( \frac{3 x}{4} - \frac{1}{4} ) ( - 9 x ^{2} + 6 x + 15 ) ^{\frac{1}{2}} + 15}{432 ( - 9 x ^{2} + 6 x + 15 ) ^{\frac{1}{2}}}

= \frac{51 x - 16 arcsin ( \frac{3 x}{4} - \frac{1}{4} ) ( - 9 x ^{2} + 6 x + 15 ) ^{\frac{1}{2}} + 15}{432 ( - 9 x ^{2} + 6 x + 15 ) ^{\frac{1}{2}}}

Füge eine Konstante zur Lösung hinzu

= \frac{51 x - 16 arcsin ( \frac{3 x}{4} - \frac{1}{4} ) ( - 9 x ^{2} + 6 x + 15 ) ^{\frac{1}{2}} + 15}{432 ( - 9 x ^{2} + 6 x + 15 ) ^{\frac{1}{2}}} + C

t an g e n t f (x) = x + 1, at x = 0

d er i v a t i v e f (x) = e^{x}

\int sin (4 x) cos (6 x) d x

\frac{d}{d x} (3 x - 4)

d er i v a t i v e - 0.01549944 x^{2} + 0.287342 x