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受欢迎的微积分 >

integral of (11x^2)/((144+x^2)^2)

解答

\int \frac{11 x ^{2}}{( 144 + x ^{2} ) ^{2}} d x

解答

\frac{11 ( x ^{2} arctan ( \frac{x}{12} ) - 12 x + 144 arctan ( \frac{x}{12} ) )}{24 ( 144 + x ^{2} )} + C

求解步骤

\int \frac{11 x ^{2}}{( 144 + x ^{2} ) ^{2}} d x

提出常数:

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

= 11 \cdot \int \frac{x ^{2}}{( 144 + x ^{2} ) ^{2}} d x

使用换元积分法

= 11 \cdot \int \frac{u ^{2}}{12 ( u ^{2} + 1 ) ^{2}} d u

提出常数:

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

= 11 \cdot \frac{1}{12} \cdot \int \frac{u ^{2}}{( u ^{2} + 1 ) ^{2}} d u

将

\frac{u ^{2}}{( u ^{2} + 1 ) ^{2}}

用部份分式展开:

\frac{1}{u ^{2} + 1} - \frac{1}{( u ^{2} + 1 ) ^{2}}

= 11 \cdot \frac{1}{12} \cdot \int \frac{1}{u ^{2} + 1} - \frac{1}{( u ^{2} + 1 ) ^{2}} d u

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

= 11 \cdot \frac{1}{12} (\int \frac{1}{u ^{2} + 1} d u - \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u)

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

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

= 11 \cdot \frac{1}{12} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

u = \frac{x}{12}

代回

= 11 \cdot \frac{1}{12} (arctan (\frac{x}{12}) - \frac{1}{2} (arctan (\frac{x}{12}) + \frac{\frac{x}{12}}{1 + ( \frac{x}{12} ) ^{2}}))

化简

11 \cdot \frac{1}{12} (arctan (\frac{x}{12}) - \frac{1}{2} (arctan (\frac{x}{12}) + \frac{\frac{x}{12}}{1 + ( \frac{x}{12} ) ^{2}})) : \frac{11 ( x ^{2} arctan ( \frac{x}{12} ) - 12 x + 144 arctan ( \frac{x}{12} ) )}{24 ( 144 + x ^{2} )}

= \frac{11 ( x ^{2} arctan ( \frac{x}{12} ) - 12 x + 144 arctan ( \frac{x}{12} ) )}{24 ( 144 + x ^{2} )}

解答补常数

= \frac{11 ( x ^{2} arctan ( \frac{x}{12} ) - 12 x + 144 arctan ( \frac{x}{12} ) )}{24 ( 144 + x ^{2} )} + C

作图

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