integral of x(x^2+1)^{100}

解答

\int x (x^{2} + 1)^{100} d x

\frac{1}{202} (x^{2} + 1)^{101} + C

求解步骤

\int x (x^{2} + 1)^{100} d x

使用换元积分法

= \int \frac{u ^{100}}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int u^{100} d u

使用幂法则

= \frac{1}{2} \cdot \frac{u ^{101}}{101}

u = x^{2} + 1

代回

= \frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{101}}{101}

化简

\frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{101}}{101} : \frac{1}{202} (x^{2} + 1)^{101}

= \frac{1}{202} (x^{2} + 1)^{101}

解答补常数

= \frac{1}{202} (x^{2} + 1)^{101} + C

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