f^'(x)=(2x^2-9x+8)/(6x+5)

Lösung

f^{^{'}} (x) = \frac{2 x ^{2} - 9 x + 8}{6 x + 5}

f (x) = \frac{x ^{2}}{6} - \frac{16}{9} x + \frac{76}{27} ln (6 x + 5) + c_{1}

Schritte zur Lösung

f^{'} (x) = \frac{2 x ^{2} - 9 x + 8}{6 x + 5}

Wenn

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

dann

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

f (x) = \int \frac{2 x ^{2} - 9 x + 8}{6 x + 5} d x

\int \frac{2 x ^{2} - 9 x + 8}{6 x + 5} d x = \frac{x ^{2}}{6} - \frac{16}{9} x + \frac{76}{27} ln (6 x + 5) + c_{1}

f (x) = \frac{x ^{2}}{6} - \frac{16}{9} x + \frac{76}{27} ln (6 x + 5) + c_{1}

\int r^{3} + 2 r d r

\frac{d}{d t} (2 cos^{2} (t))

\int x (3 x^{2} - 5)^{10} d x

\int x^{4} e^{5 x} d x

f (x) = \frac{3}{x + 5}