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

Lösung

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

\frac{4 x ^{7}}{7} + 3 x^{4} + 9 x + C

Schritte zur Lösung

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

Multipliziere aus

(2 x^{3} + 3)^{2} : 4 x^{6} + 12 x^{3} + 9

= \int 4 x^{6} + 12 x^{3} + 9 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 4 x^{6} d x + \int 12 x^{3} d x + \int 9 d x

\int 4 x^{6} d x = \frac{4 x ^{7}}{7}

\int 12 x^{3} d x = 3 x^{4}

\int 9 d x = 9 x

= \frac{4 x ^{7}}{7} + 3 x^{4} + 9 x

Füge eine Konstante zur Lösung hinzu

= \frac{4 x ^{7}}{7} + 3 x^{4} + 9 x + C

\int \frac{4 + 6 x}{6 + 4 x + 3 x ^{2}} d x

\int \frac{4 x ^{2} + 8 x}{e ^{x + 1}} d x

\int \frac{z ^{5}}{( 9 z ^{2} - 25 ) ^{\frac{3}{2}}} d z

\int \frac{x}{x ^{'}} d x

\int \frac{4 + x ^{6}}{x} d x