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受欢迎的微积分 >

integral of (4t)/(sqrt(1-16t^4))

解答

\int \frac{4 t}{1 - 16 t ^{4}} d t

解答

\frac{1}{2} arcsin (4 t^{2}) + C

求解步骤

\int \frac{4 t}{1 - 16 t ^{4}} d t

提出常数:

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

= 4 \cdot \int \frac{t}{1 - 16 t ^{4}} d t

使用换元积分法

= 4 \cdot \int \frac{1}{2 1 - 16 u ^{2}} d u

提出常数:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{1 - 16 u ^{2}} d u

使用三角换元法

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{4} d v

提出常数:

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

= 4 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \int 1 d v

常数积分:

\int a d x = a x

= 4 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot v

代回

= 4 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (4 t^{2})

化简

4 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (4 t^{2}) : \frac{1}{2} arcsin (4 t^{2})

= \frac{1}{2} arcsin (4 t^{2})

解答补常数

= \frac{1}{2} arcsin (4 t^{2}) + C

作图

流行的例子

integral of sin(t)sec^5(t)

\int sin (t) sec^{5} (t) d t

limit as x approaches 0 of (x)x

x \to 0 lim ((x) x)

integral of arccos(sqrt((1+x)/2))

\int arccos (\frac{1 + x}{2}) d x

integral of 1/(sqrt(1+u^2))

\int \frac{1}{1 + u ^{2}} d u

面积 y=x^2,y=x+6

a re a y = x^{2}, y = x + 6