integral of (2x^3+3x+1)^4(6x^2+3)

Lösung

\int (2 x^{3} + 3 x + 1)^{4} (6 x^{2} + 3) d x

\frac{32 x ^{15}}{5} + 48 x^{13} + 16 x^{12} + 144 x^{11} + 96 x^{10} + 232 x^{9} + 216 x^{8} + 234 x^{7} + 224 x^{6} + \frac{783 x ^{5}}{5} + 105 x^{4} + 56 x^{3} + 18 x^{2} + 3 x + C

Schritte zur Lösung

\int (2 x^{3} + 3 x + 1)^{4} (6 x^{2} + 3) d x

Multipliziere aus

(2 x^{3} + 3 x + 1)^{4} (6 x^{2} + 3) : 96 x^{14} + 624 x^{12} + 192 x^{11} + 1584 x^{10} + 960 x^{9} + 2088 x^{8} + 1728 x^{7} + 1638 x^{6} + 1344 x^{5} + 783 x^{4} + 420 x^{3} + 168 x^{2} + 36 x + 3

= \int 96 x^{14} + 624 x^{12} + 192 x^{11} + 1584 x^{10} + 960 x^{9} + 2088 x^{8} + 1728 x^{7} + 1638 x^{6} + 1344 x^{5} + 783 x^{4} + 420 x^{3} + 168 x^{2} + 36 x + 3 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 96 x^{14} d x + \int 624 x^{12} d x + \int 192 x^{11} d x + \int 1584 x^{10} d x + \int 960 x^{9} d x + \int 2088 x^{8} d x + \int 1728 x^{7} d x + \int 1638 x^{6} d x + \int 1344 x^{5} d x + \int 783 x^{4} d x + \int 420 x^{3} d x + \int 168 x^{2} d x + \int 36 x d x + \int 3 d x

\int 96 x^{14} d x = \frac{32 x ^{15}}{5}

\int 624 x^{12} d x = 48 x^{13}

\int 192 x^{11} d x = 16 x^{12}

\int 1584 x^{10} d x = 144 x^{11}

\int 960 x^{9} d x = 96 x^{10}

\int 2088 x^{8} d x = 232 x^{9}

\int 1728 x^{7} d x = 216 x^{8}

\int 1638 x^{6} d x = 234 x^{7}

\int 1344 x^{5} d x = 224 x^{6}

\int 783 x^{4} d x = \frac{783 x ^{5}}{5}

\int 420 x^{3} d x = 105 x^{4}

\int 168 x^{2} d x = 56 x^{3}

\int 36 x d x = 18 x^{2}

\int 3 d x = 3 x

= \frac{32 x ^{15}}{5} + 48 x^{13} + 16 x^{12} + 144 x^{11} + 96 x^{10} + 232 x^{9} + 216 x^{8} + 234 x^{7} + 224 x^{6} + \frac{783 x ^{5}}{5} + 105 x^{4} + 56 x^{3} + 18 x^{2} + 3 x

Füge eine Konstante zur Lösung hinzu

= \frac{32 x ^{15}}{5} + 48 x^{13} + 16 x^{12} + 144 x^{11} + 96 x^{10} + 232 x^{9} + 216 x^{8} + 234 x^{7} + 224 x^{6} + \frac{783 x ^{5}}{5} + 105 x^{4} + 56 x^{3} + 18 x^{2} + 3 x + C

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

(ln (1 - x^{2}))^{^{'}}

x y^{2} y^{^{'}} = y^{3} - 2 x^{3}

\frac{d}{d x} (\frac{1}{4 + 3 x})

\int_{1}^{2} x^{5} + \frac{1}{4 x ^{5}} d x