x^'= 1/(1+t^2)

解答

x^{^{'}} = \frac{1}{1 + t ^{2}}

x = arctan (t) + c_{1}

求解步骤

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

若

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

则

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

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

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

x = arctan (t) + c_{1}

t an g e n t f (x) = 4 - 2 x^{2}, at x = 5

d er i v a t i v e (\frac{- 2 x - 1}{- 4 x + 2})^{5}

\frac{\partial}{\partial y} (x^{2} + x y^{4})

\int_{1}^{x} \frac{1}{t} d t

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