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受欢迎的微积分 >

integral of (y^2)/((y^2+3)^{3/2)}

解答

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

解答

ln \frac{1}{3 ^{\frac{1}{2}}} y + \frac{1}{3} (3 + y^{2}) - \frac{y}{( 3 + y ^{2} ) ^{\frac{1}{2}}} + C

求解步骤

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

使用三角换元法

= \int tan^{2} (u) cos (u) d u

化简

tan^{2} (u) cos (u) : sin (u) tan (u)

= \int sin (u) tan (u) d u

使用三角恒等式改写

= \int \frac{1 - cos ^{2} ( u )}{cos ( u )} d u

乘开

\frac{1 - cos ^{2} ( u )}{cos ( u )} : \frac{1}{cos ( u )} - cos (u)

= \int \frac{1}{cos ( u )} - cos (u) d u

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

= \int \frac{1}{cos ( u )} d u - \int cos (u) d u

\int \frac{1}{cos ( u )} d u = ln ∣ tan (u) + sec (u) ∣

\int cos (u) d u = sin (u)

= ln ∣ tan (u) + sec (u) ∣ - sin (u)

u = arctan (\frac{1}{3 ^{\frac{1}{2}}} y)

代回

= ln tan (arctan (\frac{1}{3 ^{\frac{1}{2}}} y)) + sec (arctan (\frac{1}{3 ^{\frac{1}{2}}} y)) - sin (arctan (\frac{1}{3 ^{\frac{1}{2}}} y))

化简

ln tan (arctan (\frac{1}{3 ^{\frac{1}{2}}} y)) + sec (arctan (\frac{1}{3 ^{\frac{1}{2}}} y)) - sin (arctan (\frac{1}{3 ^{\frac{1}{2}}} y)) : ln \frac{1}{3 ^{\frac{1}{2}}} y + \frac{1}{3} (3 + y^{2}) - \frac{y}{( 3 + y ^{2} ) ^{\frac{1}{2}}}

= ln \frac{1}{3 ^{\frac{1}{2}}} y + \frac{1}{3} (3 + y^{2}) - \frac{y}{( 3 + y ^{2} ) ^{\frac{1}{2}}}

解答补常数

= ln \frac{1}{3 ^{\frac{1}{2}}} y + \frac{1}{3} (3 + y^{2}) - \frac{y}{( 3 + y ^{2} ) ^{\frac{1}{2}}} + C

作图

流行的例子

limit as h approaches 0 of (sin(9h))/(9h)

h \to 0 lim (\frac{sin ( 9 h )}{9 h})

limit as x approaches 0 of sqrt(x+9)-3

x \to 0 lim (x + 9 - 3)

derivative of (x^2/(x-8))

\frac{d}{d x} (\frac{x ^{2}}{x - 8})

integral of (x^2)/(x^2+x+6)

\int \frac{x ^{2}}{x ^{2} + x + 6} d x

tangent sqrt(x)+1/(sqrt(x^3))

t an g e n t x + \frac{1}{x ^{3}}