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受欢迎的微积分 >

integral of 5/(sqrt(16x^2+1))

解答

\int \frac{5}{16 x ^{2} + 1} d x

解答

\frac{5}{4} ln 4 x + 1 + 16 x^{2} + C

求解步骤

\int \frac{5}{16 x ^{2} + 1} d x

提出常数:

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

= 5 \cdot \int \frac{1}{16 x ^{2} + 1} d x

使用三角换元法

= 5 \cdot \int \frac{sec ( u )}{4} d u

提出常数:

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

= 5 \cdot \frac{1}{4} \cdot \int sec (u) d u

使用积分常见定则:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 5 \cdot \frac{1}{4} ln ∣ tan (u) + sec (u) ∣

u = arctan (4 x)

代回

= 5 \cdot \frac{1}{4} ln ∣ tan (arctan (4 x)) + sec (arctan (4 x)) ∣

化简

5 \cdot \frac{1}{4} ln ∣ tan (arctan (4 x)) + sec (arctan (4 x)) ∣ : \frac{5}{4} ln 4 x + 1 + 16 x^{2}

= \frac{5}{4} ln 4 x + 1 + 16 x^{2}

解答补常数

= \frac{5}{4} ln 4 x + 1 + 16 x^{2} + C

作图

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