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受欢迎的微积分 >

integral of (4sqrt(x))/(sqrt(4x+1))

解答

\int \frac{4 x}{4 x + 1} d x

解答

- \frac{1}{2} ln 2 x + 4 x + 1 + x 4 x + 1 + C

求解步骤

\int \frac{4 x}{4 x + 1} d x

提出常数:

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

= 4 \cdot \int \frac{x}{4 x + 1} d x

使用换元积分法

= 4 \cdot \int \frac{2 u ^{2}}{4 u ^{2} + 1} d u

提出常数:

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

= 4 \cdot 2 \cdot \int \frac{u ^{2}}{4 u ^{2} + 1} d u

使用三角换元法

= 4 \cdot 2 \cdot \int \frac{tan ^{2} ( v ) sec ( v )}{8} d v

提出常数:

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

= 4 \cdot 2 \cdot \frac{1}{8} \cdot \int tan^{2} (v) sec (v) d v

使用三角恒等式改写

= 4 \cdot 2 \cdot \frac{1}{8} \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

乘开

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= 4 \cdot 2 \cdot \frac{1}{8} \cdot \int - sec (v) + sec^{3} (v) d v

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

= 4 \cdot 2 \cdot \frac{1}{8} (- \int sec (v) d v + \int sec^{3} (v) d v)

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

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

= 4 \cdot 2 \cdot \frac{1}{8} (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

代回

= 4 \cdot 2 \cdot \frac{1}{8} (- ln tan (arctan (2 x)) + sec (arctan (2 x)) + \frac{1}{2} sec (arctan (2 x)) tan (arctan (2 x)) + \frac{1}{2} ln tan (arctan (2 x)) + sec (arctan (2 x)))

化简

4 \cdot 2 \cdot \frac{1}{8} (- ln tan (arctan (2 x)) + sec (arctan (2 x)) + \frac{1}{2} sec (arctan (2 x)) tan (arctan (2 x)) + \frac{1}{2} ln tan (arctan (2 x)) + sec (arctan (2 x))) : - \frac{1}{2} ln 2 x + 4 x + 1 + x 4 x + 1

= - \frac{1}{2} ln 2 x + 4 x + 1 + x 4 x + 1

解答补常数

= - \frac{1}{2} ln 2 x + 4 x + 1 + x 4 x + 1 + C

作图

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