integral of-1/4 x^3+3/2 x^2+1

Lösung

\int - \frac{1}{4} x^{3} + \frac{3}{2} x^{2} + 1 d x

- \frac{x ^{4}}{16} + \frac{x ^{3}}{2} + x + C

Schritte zur Lösung

\int - \frac{1}{4} x^{3} + \frac{3}{2} x^{2} + 1 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 \frac{1}{4} x^{3} d x + \int \frac{3}{2} x^{2} d x + \int 1 d x

\int \frac{1}{4} x^{3} d x = \frac{x ^{4}}{16}

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

\int 1 d x = x

= - \frac{x ^{4}}{16} + \frac{x ^{3}}{2} + x

Füge eine Konstante zur Lösung hinzu

= - \frac{x ^{4}}{16} + \frac{x ^{3}}{2} + x + C

\int \frac{5 x}{1 + x ^{4}} d x

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

\int \frac{ln ( x + 1 )}{x} d x

\int (8 x^{9}) d x

\int \frac{3 x}{( 1 + 9 x ^{2} ) ^{4}} d x