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受欢迎的微积分 >

integral of 8/((16x^2+1)^2)

解答

\int \frac{8}{( 16 x ^{2} + 1 ) ^{2}} d x

解答

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

求解步骤

\int \frac{8}{( 16 x ^{2} + 1 ) ^{2}} d x

提出常数:

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

= 8 \cdot \int \frac{1}{( 16 x ^{2} + 1 ) ^{2}} d x

使用换元积分法

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

提出常数:

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

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

使用三角换元法

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

使用三角恒等式改写

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

提出常数:

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

= 8 \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \int 1 + cos (2 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

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

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 8 \cdot \frac{1}{4} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

代回

= 8 \cdot \frac{1}{4} \cdot \frac{1}{2} (arctan (4 x) + \frac{1}{2} sin (2 arctan (4 x)))

化简

8 \cdot \frac{1}{4} \cdot \frac{1}{2} (arctan (4 x) + \frac{1}{2} sin (2 arctan (4 x))) : arctan (4 x) + \frac{1}{2} sin (2 arctan (4 x))

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

解答补常数

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

作图

流行的例子

(\partial)/(\partial y)(e^{xyz^6}yz^6)

\frac{\partial}{\partial y} (e^{x y z^{6}} y z^{6})

麦克劳林 sqrt(1-x)

ma c l a u r in 1 - x

y^{''}+8y^'+16y=1521e^{2t}cos(9t)

y^{^{''}} + 8 y^{^{'}} + 16 y = 1521 e^{2 t} cos (9 t)

laplacetransform cos^2(1t)

l a pl a ce t r an s f or m cos^{2} (1 t)

tangent f(x)= 1/(x-1),\at x=5

t an g e n t f (x) = \frac{1}{x - 1}, at x = 5