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受欢迎的微积分 >

integral of 4x^7sqrt(x^4+1)

解答

\int 4 x^{7} x^{4} + 1 d x

解答

\frac{2}{5} (x^{4} + 1)^{\frac{5}{2}} - \frac{2}{3} (x^{4} + 1)^{\frac{3}{2}} + C

求解步骤

\int 4 x^{7} x^{4} + 1 d x

提出常数:

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

= 4 \cdot \int x^{7} x^{4} + 1 d x

使用换元积分法

= 4 \cdot \int \frac{u ^{\frac{1}{7}} u ^{\frac{4}{7}} + 1}{7} d u

提出常数:

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

= 4 \cdot \frac{1}{7} \cdot \int u^{\frac{1}{7}} u^{\frac{4}{7}} + 1 d u

使用换元积分法

= 4 \cdot \frac{1}{7} \cdot \int \frac{7 ( v - 1 ) v}{4} d v

提出常数:

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

= 4 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \int (v - 1) v d v

乘开

(v - 1) v : v^{\frac{3}{2}} - v

= 4 \cdot \frac{1}{7} \cdot \frac{7}{4} \cdot \int v^{\frac{3}{2}} - v d v

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

= 4 \cdot \frac{1}{7} \cdot \frac{7}{4} (\int v^{\frac{3}{2}} d v - \int v d v)

\int v^{\frac{3}{2}} d v = \frac{2}{5} v^{\frac{5}{2}}

\int v d v = \frac{2}{3} v^{\frac{3}{2}}

= 4 \cdot \frac{1}{7} \cdot \frac{7}{4} (\frac{2}{5} v^{\frac{5}{2}} - \frac{2}{3} v^{\frac{3}{2}})

代回

= 4 \cdot \frac{1}{7} \cdot \frac{7}{4} (\frac{2}{5} ((x^{7})^{\frac{4}{7}} + 1)^{\frac{5}{2}} - \frac{2}{3} ((x^{7})^{\frac{4}{7}} + 1)^{\frac{3}{2}})

化简

4 \cdot \frac{1}{7} \cdot \frac{7}{4} (\frac{2}{5} ((x^{7})^{\frac{4}{7}} + 1)^{\frac{5}{2}} - \frac{2}{3} ((x^{7})^{\frac{4}{7}} + 1)^{\frac{3}{2}}) : \frac{2}{5} (x^{4} + 1)^{\frac{5}{2}} - \frac{2}{3} (x^{4} + 1)^{\frac{3}{2}}

= \frac{2}{5} (x^{4} + 1)^{\frac{5}{2}} - \frac{2}{3} (x^{4} + 1)^{\frac{3}{2}}

解答补常数

= \frac{2}{5} (x^{4} + 1)^{\frac{5}{2}} - \frac{2}{3} (x^{4} + 1)^{\frac{3}{2}} + C

作图

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