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受欢迎的微积分 >

integral of 24arctan(sqrt(x))

解答

\int 24 arctan (x) d x

解答

48 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x)) + C

求解步骤

\int 24 arctan (x) d x

提出常数:

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

= 24 \cdot \int arctan (x) d x

使用换元积分法

= 24 \cdot \int arctan (u) \cdot 2 u d u

提出常数:

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

= 24 \cdot 2 \cdot \int arctan (u) u d u

使用分布积分法

= 24 \cdot 2 (\frac{1}{2} u^{2} arctan (u) - \int \frac{u ^{2}}{2 ( u ^{2} + 1 )} d u)

\int \frac{u ^{2}}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} (- arctan (u) + u)

= 24 \cdot 2 (\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u))

u = x

代回

= 24 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x))

化简

24 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)) : 48 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

= 48 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

解答补常数

= 48 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x)) + C

作图

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