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

Lösung

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

sin (3) \frac{π}{2} C (\frac{2}{π} x) + cos (3) \frac{π}{2} § (\frac{2}{π} x) - \frac{x ^{4}}{4} + 2 x^{2} + C

Schritte zur Lösung

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

\int sin (x^{2} + 3) d x = sin (3) \frac{π}{2} C (\frac{2}{π} x) + cos (3) \frac{π}{2} § (\frac{2}{π} x)

\int x^{3} d x = \frac{x ^{4}}{4}

\int 4 x d x = 2 x^{2}

= sin (3) \frac{π}{2} C (\frac{2}{π} x) + cos (3) \frac{π}{2} § (\frac{2}{π} x) - \frac{x ^{4}}{4} + 2 x^{2}

Füge eine Konstante zur Lösung hinzu

= sin (3) \frac{π}{2} C (\frac{2}{π} x) + cos (3) \frac{π}{2} § (\frac{2}{π} x) - \frac{x ^{4}}{4} + 2 x^{2} + C

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

\int \frac{( 6 + sin ( x ) )}{cos ( x )} d x

\int \frac{3}{x ^{4}} \frac{d}{1} d x

\int \frac{4}{x ^{4} x ^{2} - 8} d x

\int (- 3 x y^{2} - 2 y cos (x)) d y