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

解答

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

\frac{1}{4} (2 x 4 x + 1 + ln 2 x + 4 x + 1) + C

求解步骤

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

化简

\frac{4 x + 1}{4 x} : \frac{4 x + 1}{2 x}

= \int \frac{4 x + 1}{2 x} d x

提出常数:

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

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

使用分布积分法

= \frac{1}{2} (2 x 4 x + 1 - \int \frac{4 x}{4 x + 1} d x)

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

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

化简

\frac{1}{2} (2 x 4 x + 1 - (- \frac{1}{2} ln 2 x + 4 x + 1 + x 4 x + 1)) : \frac{1}{4} (2 x 4 x + 1 + ln 2 x + 4 x + 1)

= \frac{1}{4} (2 x 4 x + 1 + ln 2 x + 4 x + 1)

解答补常数

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

n = 1 \sum \infty \frac{1}{n ^{2}} - \frac{1}{n}

x \to 5 lim (x^{2} - 25)

\int \frac{2}{( x ^{2} - 3 x + 2 )} d x

d er i v a t i v e f (t) = \frac{t}{t ^{2} + 7}

\frac{d}{d x} (e^{x} (6 x^{2} - 12 x + 12))