integral of 8x^3(1+x^4)^{10}

解答

\int 8 x^{3} (1 + x^{4})^{10} d x

8 (\frac{x ^{4}}{4} + \frac{5 x ^{8}}{4} + \frac{15 x ^{12}}{4} + \frac{15 x ^{16}}{2} + \frac{21 x ^{20}}{2} + \frac{21 x ^{24}}{2} + \frac{15 x ^{28}}{2} + \frac{15 x ^{32}}{4} + \frac{5 x ^{36}}{4} + \frac{x ^{40}}{4} + \frac{x ^{44}}{44}) + C

求解步骤

\int 8 x^{3} (1 + x^{4})^{10} d x

提出常数:

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

= 8 \cdot \int x^{3} (1 + x^{4})^{10} d x

乘开

x^{3} (1 + x^{4})^{10} : x^{3} + 10 x^{7} + 45 x^{11} + 120 x^{15} + 210 x^{19} + 252 x^{23} + 210 x^{27} + 120 x^{31} + 45 x^{35} + 10 x^{39} + x^{43}

= 8 \cdot \int x^{3} + 10 x^{7} + 45 x^{11} + 120 x^{15} + 210 x^{19} + 252 x^{23} + 210 x^{27} + 120 x^{31} + 45 x^{35} + 10 x^{39} + x^{43} d x

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

= 8 (\int x^{3} d x + \int 10 x^{7} d x + \int 45 x^{11} d x + \int 120 x^{15} d x + \int 210 x^{19} d x + \int 252 x^{23} d x + \int 210 x^{27} d x + \int 120 x^{31} d x + \int 45 x^{35} d x + \int 10 x^{39} d x + \int x^{43} d x)

\int x^{3} d x = \frac{x ^{4}}{4}

\int 10 x^{7} d x = \frac{5 x ^{8}}{4}

\int 45 x^{11} d x = \frac{15 x ^{12}}{4}

\int 120 x^{15} d x = \frac{15 x ^{16}}{2}

\int 210 x^{19} d x = \frac{21 x ^{20}}{2}

\int 252 x^{23} d x = \frac{21 x ^{24}}{2}

\int 210 x^{27} d x = \frac{15 x ^{28}}{2}

\int 120 x^{31} d x = \frac{15 x ^{32}}{4}

\int 45 x^{35} d x = \frac{5 x ^{36}}{4}

\int 10 x^{39} d x = \frac{x ^{40}}{4}

\int x^{43} d x = \frac{x ^{44}}{44}

= 8 (\frac{x ^{4}}{4} + \frac{5 x ^{8}}{4} + \frac{15 x ^{12}}{4} + \frac{15 x ^{16}}{2} + \frac{21 x ^{20}}{2} + \frac{21 x ^{24}}{2} + \frac{15 x ^{28}}{2} + \frac{15 x ^{32}}{4} + \frac{5 x ^{36}}{4} + \frac{x ^{40}}{4} + \frac{x ^{44}}{44})

解答补常数

= 8 (\frac{x ^{4}}{4} + \frac{5 x ^{8}}{4} + \frac{15 x ^{12}}{4} + \frac{15 x ^{16}}{2} + \frac{21 x ^{20}}{2} + \frac{21 x ^{24}}{2} + \frac{15 x ^{28}}{2} + \frac{15 x ^{32}}{4} + \frac{5 x ^{36}}{4} + \frac{x ^{40}}{4} + \frac{x ^{44}}{44}) + C