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受欢迎的微积分 >

integral of x*ln(sqrt(x^2+1))

解答

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

解答

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

求解步骤

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

化简

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

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

提出常数:

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

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

使用换元积分法

= \frac{1}{2} \cdot \int \frac{ln ( u )}{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 \frac{1}{2} \cdot \int ln (u) d u

使用分布积分法

= \frac{1}{2} \cdot \frac{1}{2} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{2} \cdot \frac{1}{2} (u ln (u) - u)

u = x^{2} + 1

代回

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

化简

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

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

解答补常数

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

作图

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