dx=4(x^2+1)dt

Lösung

d x = 4 (x^{2} + 1) d t

t (x) = \frac{1}{4} arctan (x) + c_{1}

Schritte zur Lösung

d x = 4 (x^{2} + 1) d t (x)

Lass

t (x)

die abhängige Variable sein. Teile durch

d x

1 = 4 (x^{2} + 1) \frac{d t ( x )}{d x}

Ersetze

\frac{d t ( x )}{d x}

mit

t^{'} (x)

1 = 4 (x^{2} + 1) t^{^{'}} (x)

Isoliere

t^{'} (x) : t^{'} (x) = \frac{1}{4 ( x ^{2} + 1 )}

Wenn

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

dann

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

t (x) = \int \frac{1}{4 ( x ^{2} + 1 )} d x

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

t (x) = \frac{1}{4} arctan (x) + c_{1}

in v erse l a pl a ce \frac{1}{( s + 5 )}

in v erse l a pl a ce \frac{1}{5 s - 2}

\int \frac{x ^{2}}{2 x} d x

\int c^{x} d x

x \to - 2 lim (4 x^{2} + 3 x - 1)