integral of (x^2+1)^32x

解答

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

\frac{x ^{8}}{4} + x^{6} + \frac{3 x ^{4}}{2} + x^{2} + C

求解步骤

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

提出常数:

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

= 2 \cdot \int (x^{2} + 1)^{3} x d x

乘开

(x^{2} + 1)^{3} x : x^{7} + 3 x^{5} + 3 x^{3} + x

= 2 \cdot \int x^{7} + 3 x^{5} + 3 x^{3} + x 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

= 2 (\int x^{7} d x + \int 3 x^{5} d x + \int 3 x^{3} d x + \int x d x)

\int x^{7} d x = \frac{x ^{8}}{8}

\int 3 x^{5} d x = \frac{x ^{6}}{2}

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

\int x d x = \frac{x ^{2}}{2}

= 2 (\frac{x ^{8}}{8} + \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} + \frac{x ^{2}}{2})

化简

2 (\frac{x ^{8}}{8} + \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} + \frac{x ^{2}}{2}) : \frac{x ^{8}}{4} + x^{6} + \frac{3 x ^{4}}{2} + x^{2}

= \frac{x ^{8}}{4} + x^{6} + \frac{3 x ^{4}}{2} + x^{2}

解答补常数

= \frac{x ^{8}}{4} + x^{6} + \frac{3 x ^{4}}{2} + x^{2} + C

\int (5 - 4 x) e^{6 - 5 x} d x

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