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受欢迎的微积分 >

integral of (6400q)/(16q^2+3200)

解答

\int \frac{6400 q}{16 q ^{2} + 3200} d q

解答

200 ln 16 q^{2} + 3200 + C

求解步骤

\int \frac{6400 q}{16 q ^{2} + 3200} d q

提出常数:

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

= 6400 \cdot \int \frac{q}{16 q ^{2} + 3200} d q

使用换元积分法

= 6400 \cdot \int \frac{1}{32 u} d u

提出常数:

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

= 6400 \cdot \frac{1}{32} \cdot \int \frac{1}{u} d u

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6400 \cdot \frac{1}{32} ln ∣ u ∣

u = 16 q^{2} + 3200

代回

= 6400 \cdot \frac{1}{32} ln 16 q^{2} + 3200

化简

6400 \cdot \frac{1}{32} ln 16 q^{2} + 3200 : 200 ln 16 q^{2} + 3200

= 200 ln 16 q^{2} + 3200

解答补常数

= 200 ln 16 q^{2} + 3200 + C

作图

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