integral of (8x+3sin(x))^2

Lösung

\int (8 x + 3 sin (x))^{2} d x

\frac{64 x ^{3}}{3} + 48 (- x cos (x) + sin (x)) + \frac{9}{2} (x - \frac{1}{2} sin (2 x)) + C

Schritte zur Lösung

\int (8 x + 3 sin (x))^{2} d x

Multipliziere aus

(8 x + 3 sin (x))^{2} : 64 x^{2} + 48 x sin (x) + 9 sin^{2} (x)

= \int 64 x^{2} + 48 x sin (x) + 9 sin^{2} (x) d x

Wende die Summenregel an:

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

= \int 64 x^{2} d x + \int 48 x sin (x) d x + \int 9 sin^{2} (x) d x

\int 64 x^{2} d x = \frac{64 x ^{3}}{3}

\int 48 x sin (x) d x = 48 (- x cos (x) + sin (x))

\int 9 sin^{2} (x) d x = \frac{9}{2} (x - \frac{1}{2} sin (2 x))

= \frac{64 x ^{3}}{3} + 48 (- x cos (x) + sin (x)) + \frac{9}{2} (x - \frac{1}{2} sin (2 x))

Füge eine Konstante zur Lösung hinzu

= \frac{64 x ^{3}}{3} + 48 (- x cos (x) + sin (x)) + \frac{9}{2} (x - \frac{1}{2} sin (2 x)) + C

\int (\frac{1}{4 - 25 x ^{2}}) d x

d er i v a t i v e \frac{14 x + 4 x ^{2}}{( 7 + 4 x ) ^{2}}

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

\frac{d y}{d x} = \frac{( 2 x ( y - 1 ) )}{( x ^{2} + 3 )}

t an g e n t y = 2 x^{3} - x^{2} + 2 (1.3)