f^'(x)=2^{4x}+4x^2

Lösung

f^{^{'}} (x) = 2^{4 x} + 4 x^{2}

f (x) = \frac{2 ^{4 x - 2}}{ln ( 2 )} + \frac{4 x ^{3}}{3} + c_{1}

Schritte zur Lösung

f^{'} (x) = 2^{4 x} + 4 x^{2}

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

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

\int 2^{4 x} + 4 x^{2} d x = \frac{2 ^{4 x - 2}}{ln ( 2 )} + \frac{4 x ^{3}}{3} + c_{1}

f (x) = \frac{2 ^{4 x - 2}}{ln ( 2 )} + \frac{4 x ^{3}}{3} + c_{1}

\int (1 - 2 x)^{9} d x

d er i v a t i v e e^{t} \cdot cos (2 t)

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

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

x \frac{d y}{d x} + y = y^{- 2}