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受欢迎的微积分 >

integral of (xln(x^2+1))/(x^2+1)

解答

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

解答

\frac{1}{4} ln^{2} (x^{2} + 1) + C

求解步骤

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

使用换元积分法

= \int \frac{ln ( u )}{2 u} 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 \frac{ln ( u )}{u} d u

使用换元积分法

= \frac{1}{2} \cdot \int v d v

使用幂法则

= \frac{1}{2} \cdot \frac{v ^{2}}{2}

代回

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

化简

\frac{1}{2} \cdot \frac{ln ^{2} ( x ^{2} + 1 )}{2} : \frac{1}{4} ln^{2} (x^{2} + 1)

= \frac{1}{4} ln^{2} (x^{2} + 1)

解答补常数

= \frac{1}{4} ln^{2} (x^{2} + 1) + C

作图

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\int sec (a) x d x

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\frac{d}{d x} ((r (x) (x)^{2} - x^{2})^{\frac{1}{2}})

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\frac{d}{d x} (sin^{2} (b x) + x^{2} + 1)