integral of (t^2+2t+1)^4

解答

\int (t^{2} + 2 t + 1)^{4} d t

\frac{t ^{9}}{9} + t^{8} + 4 t^{7} + \frac{28 t ^{6}}{3} + 14 t^{5} + 14 t^{4} + \frac{28 t ^{3}}{3} + 4 t^{2} + t + C

求解步骤

\int (t^{2} + 2 t + 1)^{4} d t

乘开

(t^{2} + 2 t + 1)^{4} : t^{8} + 8 t^{7} + 28 t^{6} + 56 t^{5} + 70 t^{4} + 56 t^{3} + 28 t^{2} + 8 t + 1

= \int t^{8} + 8 t^{7} + 28 t^{6} + 56 t^{5} + 70 t^{4} + 56 t^{3} + 28 t^{2} + 8 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^{8} d t + \int 8 t^{7} d t + \int 28 t^{6} d t + \int 56 t^{5} d t + \int 70 t^{4} d t + \int 56 t^{3} d t + \int 28 t^{2} d t + \int 8 t d t + \int 1 d t

\int t^{8} d t = \frac{t ^{9}}{9}

\int 8 t^{7} d t = t^{8}

\int 28 t^{6} d t = 4 t^{7}

\int 56 t^{5} d t = \frac{28 t ^{6}}{3}

\int 70 t^{4} d t = 14 t^{5}

\int 56 t^{3} d t = 14 t^{4}

\int 28 t^{2} d t = \frac{28 t ^{3}}{3}

\int 8 t d t = 4 t^{2}

\int 1 d t = t

= \frac{t ^{9}}{9} + t^{8} + 4 t^{7} + \frac{28 t ^{6}}{3} + 14 t^{5} + 14 t^{4} + \frac{28 t ^{3}}{3} + 4 t^{2} + t

解答补常数

= \frac{t ^{9}}{9} + t^{8} + 4 t^{7} + \frac{28 t ^{6}}{3} + 14 t^{5} + 14 t^{4} + \frac{28 t ^{3}}{3} + 4 t^{2} + t + C

25 y^{^{''}} + 3 y = 0

\int_{0}^{4} 2 π x (8 - x^{\frac{3}{2}}) d x

\int_{2}^{3} \frac{33}{3 - x} d x

\frac{d x}{d t} = e^{x} \cdot sin (t)

ma c l a u r in sin (\frac{x π}{4})