integral of (1+y^3)^2

解答

\int (1 + y^{3})^{2} d y

y + \frac{y ^{4}}{2} + \frac{y ^{7}}{7} + C

求解步骤

\int (1 + y^{3})^{2} d y

乘开

(1 + y^{3})^{2} : 1 + 2 y^{3} + y^{6}

= \int 1 + 2 y^{3} + y^{6} d y

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 1 d y + \int 2 y^{3} d y + \int y^{6} d y

\int 1 d y = y

\int 2 y^{3} d y = \frac{y ^{4}}{2}

\int y^{6} d y = \frac{y ^{7}}{7}

= y + \frac{y ^{4}}{2} + \frac{y ^{7}}{7}

解答补常数

= y + \frac{y ^{4}}{2} + \frac{y ^{7}}{7} + C

d er i v a t i v e y = \frac{3 + x f ( x )}{x}

x \to - \infty lim (- 1)

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