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受欢迎的微积分 >

integral of sqrt(4+y^2)

解答

\int 4 + y^{2} d y

解答

4 \frac{1}{8} y 4 + y^{2} + \frac{1}{2} ln \frac{y + 4 + y ^{2}}{2} + C

求解步骤

\int 4 + y^{2} d y

使用三角换元法

= \int 4 sec^{3} (u) d u

提出常数:

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

= 4 \cdot \int sec^{3} (u) d u

使用降次积分法

= 4 (\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) ∣

= 4 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (\frac{1}{2} y)

代回

= 4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} y ) ) sin ( arctan ( \frac{1}{2} y ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} y)) + sec (arctan (\frac{1}{2} y)))

化简

4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} y ) ) sin ( arctan ( \frac{1}{2} y ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} y)) + sec (arctan (\frac{1}{2} y))) : 4 \frac{1}{8} y 4 + y^{2} + \frac{1}{2} ln \frac{y + 4 + y ^{2}}{2}

= 4 \frac{1}{8} y 4 + y^{2} + \frac{1}{2} ln \frac{y + 4 + y ^{2}}{2}

解答补常数

= 4 \frac{1}{8} y 4 + y^{2} + \frac{1}{2} ln \frac{y + 4 + y ^{2}}{2} + C

作图

流行的例子

integral of (14)/((x-1)^2(x-2)^2)

\int \frac{14}{( x - 1 ) ^{2} ( x - 2 ) ^{2}} d x

f(x)=arcsin(4x)

f (x) = arcsin (4 x)

limit as x approaches infinity of arctan(x^2-x^5)

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

tangent y=e^xln(x),(1,0)

t an g e n t y = e^{x} ln (x), (1, 0)

(\partial)/(\partial y)(e^xy^2)

\frac{\partial}{\partial y} (e^{x} y^{2})