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受欢迎的微积分 >

integral of 6/(5zsqrt(4z^4-9))

解答

\int \frac{6}{5 z 4 z ^{4} - 9} d z

解答

\frac{1}{5} arctan (\frac{4 z ^{4} - 9}{3}) + C

求解步骤

\int \frac{6}{5 z 4 z ^{4} - 9} d z

提出常数:

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

= \frac{6}{5} \cdot \int \frac{1}{z 4 z ^{4} - 9} d z

使用换元积分法

= \frac{6}{5} \cdot \int \frac{1}{4 u 4 u - 9} d u

提出常数:

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

= \frac{6}{5} \cdot \frac{1}{4} \cdot \int \frac{1}{u 4 u - 9} d u

使用换元积分法

= \frac{6}{5} \cdot \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 9} d v

提出常数:

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

= \frac{6}{5} \cdot \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} + 9} d v

使用换元积分法

= \frac{6}{5} \cdot \frac{1}{4} \cdot 2 \cdot \int \frac{1}{3 ( w ^{2} + 1 )} d w

提出常数:

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

= \frac{6}{5} \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{3} \cdot \int \frac{1}{w ^{2} + 1} d w

使用积分常见定则:

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

= \frac{6}{5} \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{3} arctan (w)

代回

= \frac{6}{5} \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{3} arctan (\frac{4 z ^{4} - 9}{3})

化简

\frac{6}{5} \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{3} arctan (\frac{4 z ^{4} - 9}{3}) : \frac{1}{5} arctan (\frac{4 z ^{4} - 9}{3})

= \frac{1}{5} arctan (\frac{4 z ^{4} - 9}{3})

解答补常数

= \frac{1}{5} arctan (\frac{4 z ^{4} - 9}{3}) + C

作图

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