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受欢迎的微积分 >

integral of 9/(25+16r^2)

解答

\int \frac{9}{25 + 16 r ^{2}} d r

解答

\frac{9}{20} arctan (\frac{4}{5} r) + C

求解步骤

\int \frac{9}{25 + 16 r ^{2}} d r

提出常数:

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

= 9 \cdot \int \frac{1}{25 + 16 r ^{2}} d r

使用换元积分法

= 9 \cdot \int \frac{1}{20 ( u ^{2} + 1 )} d u

提出常数:

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

= 9 \cdot \frac{1}{20} \cdot \int \frac{1}{u ^{2} + 1} d u

使用积分常见定则:

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

= 9 \cdot \frac{1}{20} arctan (u)

u = \frac{4}{5} r

代回

= 9 \cdot \frac{1}{20} arctan (\frac{4}{5} r)

化简

9 \cdot \frac{1}{20} arctan (\frac{4}{5} r) : \frac{9}{20} arctan (\frac{4}{5} r)

= \frac{9}{20} arctan (\frac{4}{5} r)

解答补常数

= \frac{9}{20} arctan (\frac{4}{5} r) + C

作图

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