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

解答

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

- arctan (y) + C

求解步骤

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

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

使用积分常见定则:

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

= - arctan (y)

解答补常数

= - arctan (y) + C

y^{^{''}} + 2 y = sin (2 t), y (0) = 0, y^{^{'}} (0) = 0

\int \frac{1}{sec ( x ) + tan ( x )} d x

in v erse l a pl a ce \frac{1}{s ^{3} + 8 s ^{2} + 12 s}

d er i v a t i v e f (x) = 4 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 4 x + 1

n = 0 \sum \infty 2 \cdot 0. 2^{n}