integral of (4t-8)(t^2-4t+1)^5

解答

\int (4 t - 8) (t^{2} - 4 t + 1)^{5} d t

\frac{t ^{12}}{3} - 8 t^{11} + 82 t^{10} - \frac{1400 t ^{9}}{3} + 1605 t^{8} - 3408 t^{7} + 4412 t^{6} - 3408 t^{5} + 1605 t^{4} - \frac{1400 t ^{3}}{3} + 82 t^{2} - 8 t + C

求解步骤

\int (4 t - 8) (t^{2} - 4 t + 1)^{5} d t

乘开

(4 t - 8) (t^{2} - 4 t + 1)^{5} : 4 t^{11} - 88 t^{10} + 820 t^{9} - 4200 t^{8} + 12840 t^{7} - 23856 t^{6} + 26472 t^{5} - 17040 t^{4} + 6420 t^{3} - 1400 t^{2} + 164 t - 8

= \int 4 t^{11} - 88 t^{10} + 820 t^{9} - 4200 t^{8} + 12840 t^{7} - 23856 t^{6} + 26472 t^{5} - 17040 t^{4} + 6420 t^{3} - 1400 t^{2} + 164 t - 8 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 4 t^{11} d t - \int 88 t^{10} d t + \int 820 t^{9} d t - \int 4200 t^{8} d t + \int 12840 t^{7} d t - \int 23856 t^{6} d t + \int 26472 t^{5} d t - \int 17040 t^{4} d t + \int 6420 t^{3} d t - \int 1400 t^{2} d t + \int 164 t d t - \int 8 d t

\int 4 t^{11} d t = \frac{t ^{12}}{3}

\int 88 t^{10} d t = 8 t^{11}

\int 820 t^{9} d t = 82 t^{10}

\int 4200 t^{8} d t = \frac{1400 t ^{9}}{3}

\int 12840 t^{7} d t = 1605 t^{8}

\int 23856 t^{6} d t = 3408 t^{7}

\int 26472 t^{5} d t = 4412 t^{6}

\int 17040 t^{4} d t = 3408 t^{5}

\int 6420 t^{3} d t = 1605 t^{4}

\int 1400 t^{2} d t = \frac{1400 t ^{3}}{3}

\int 164 t d t = 82 t^{2}

\int 8 d t = 8 t

= \frac{t ^{12}}{3} - 8 t^{11} + 82 t^{10} - \frac{1400 t ^{9}}{3} + 1605 t^{8} - 3408 t^{7} + 4412 t^{6} - 3408 t^{5} + 1605 t^{4} - \frac{1400 t ^{3}}{3} + 82 t^{2} - 8 t

解答补常数

= \frac{t ^{12}}{3} - 8 t^{11} + 82 t^{10} - \frac{1400 t ^{9}}{3} + 1605 t^{8} - 3408 t^{7} + 4412 t^{6} - 3408 t^{5} + 1605 t^{4} - \frac{1400 t ^{3}}{3} + 82 t^{2} - 8 t + C

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