integral of t-1

解答

\int t - 1 d t

\frac{t ^{2}}{2} - t + C

求解步骤

\int t - 1 d t

使用积分加法定则:

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

= \int t d t - \int 1 d t

\int t d t = \frac{t ^{2}}{2}

\int 1 d t = t

= \frac{t ^{2}}{2} - t

解答补常数

= \frac{t ^{2}}{2} - t + C

\int \frac{sin ( x )}{5 cos ( x ) + 10} d x

\frac{\partial}{\partial x} (\frac{x}{x - y})

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

\frac{\partial}{\partial x} (x^{2} - ln (x y))

\frac{d}{d x} (π csc (π + \frac{x ^{6}}{3}))