de

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Beliebt Rechnen >

integral of (x+9)/(sqrt(16-(x-4)^2))

Lösung

\int \frac{x + 9}{16 - ( x - 4 ) ^{2}} d x

Lösung

- - x^{2} + 8 x + 13 arcsin (\frac{1}{4} x - 1) + C

Schritte zur Lösung

\int \frac{x + 9}{16 - ( x - 4 ) ^{2}} d x

Wende U-Substitution an

= \int \frac{u + 13}{16 - u ^{2}} d u

Trigonometrische Substitution anwenden

= \int 4 sin (v) + 13 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

= \int 4 sin (v) d v + \int 13 d v

\int 4 sin (v) d v = - 4 cos (v)

\int 13 d v = 13 v

= - 4 cos (v) + 13 v

Ersetze zurück

= - 4 cos (arcsin (\frac{1}{4} (x - 4))) + 13 arcsin (\frac{1}{4} (x - 4))

Vereinfache

- 4 cos (arcsin (\frac{1}{4} (x - 4))) + 13 arcsin (\frac{1}{4} (x - 4)) : - - x^{2} + 8 x + 13 arcsin (\frac{1}{4} x - 1)

= - - x^{2} + 8 x + 13 arcsin (\frac{1}{4} x - 1)

Füge eine Konstante zur Lösung hinzu

= - - x^{2} + 8 x + 13 arcsin (\frac{1}{4} x - 1) + C

Graph

Beliebte Beispiele

integral of (sin(x))^{11}(cos(x))^5

\int (sin (x))^{11} (cos (x))^{5} d x

y^'=y^3-6x^2y^3,y(1)=-1/2

y^{^{'}} = y^{3} - 6 x^{2} y^{3}, y (1) = - \frac{1}{2}

limit as x approaches 1 of (x^2+1)/(x+1)

x \to 1 lim (\frac{x ^{2} + 1}{x + 1})

derivative f(x)=80640(2x+5)^2

d er i v a t i v e f (x) = 80640 (2 x + 5)^{2}

tangent f(x)=x^2-x

t an g e n t f (x) = x^{2} - x