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受欢迎的微积分 >

integral of sqrt(4x^2+1)

解答

\int 4 x^{2} + 1 d x

解答

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

求解步骤

\int 4 x^{2} + 1 d x

使用三角换元法

= \int \frac{1}{2} sec^{3} (u) d u

提出常数:

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

= \frac{1}{2} \cdot \int sec^{3} (u) d u

使用降次积分法

= \frac{1}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

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

= \frac{1}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (2 x)

代回

= \frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 x ) ) sin ( arctan ( 2 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 x)) + sec (arctan (2 x)) ∣)

化简

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

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

解答补常数

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

作图

流行的例子

derivative 3/(5x+2)

d er i v a t i v e \frac{3}{5 x + 2}

f (x) = 6 x + 5

(\partial)/(\partial x)(((4x-4y))/(4x+3y))

\frac{\partial}{\partial x} (\frac{( 4 x - 4 y )}{4 x + 3 y})

tangent y=(x^3-9x)^8,(-3,0)

t an g e n t y = (x^{3} - 9 x)^{8}, (- 3, 0)

f (x) = \frac{1}{x - 3}