integral of (1-t^2)/(1+t^2)

解答

\int \frac{1 - t ^{2}}{1 + t ^{2}} d t

2 arctan (t) - t + C

求解步骤

\int \frac{1 - t ^{2}}{1 + t ^{2}} d t

\frac{1 - t ^{2}}{1 + t ^{2}} = \frac{2}{1 + t ^{2}} - 1

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

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

\int 1 d t = t

= 2 arctan (t) - t

解答补常数

= 2 arctan (t) - t + C

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

\frac{\partial}{\partial x} (3 y^{3} - 4 y x^{2})

\int_{0}^{7} f (x) d x

θ \to 0 lim (\frac{sin ( 4 θ )}{θ})

d er i v a t i v e f (x) = 8 x^{2} + 4 e^{- 0.5 x}