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

Lösung

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

\frac{16 x ^{5}}{5} + 8 x^{3} + 9 x + C

Schritte zur Lösung

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

Multipliziere aus

(4 x^{2} + 3)^{2} : 16 x^{4} + 24 x^{2} + 9

= \int 16 x^{4} + 24 x^{2} + 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 16 x^{4} d x + \int 24 x^{2} d x + \int 9 d x

\int 16 x^{4} d x = \frac{16 x ^{5}}{5}

\int 24 x^{2} d x = 8 x^{3}

\int 9 d x = 9 x

= \frac{16 x ^{5}}{5} + 8 x^{3} + 9 x

Füge eine Konstante zur Lösung hinzu

= \frac{16 x ^{5}}{5} + 8 x^{3} + 9 x + C

\frac{\partial}{\partial x} (5 y - 2 x)

\int x^{3} ln (1 + x) d x

d er i v a t i v e 5 cos^{2} (x) + 5 sin (x)

f (x) = arcsin (x)

in v erse l a pl a ce \frac{2}{4 s ^{2} - 8 s}