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受欢迎的微积分 >

integral of t/(1+(3t)^4)

解答

\int \frac{t}{1 + ( 3 t ) ^{4}} d t

解答

\frac{1}{18} arctan (9 t^{2}) + C

求解步骤

\int \frac{t}{1 + ( 3 t ) ^{4}} d t

化简

\frac{t}{1 + ( 3 t ) ^{4}} : \frac{t}{1 + 81 t ^{4}}

= \int \frac{t}{1 + 81 t ^{4}} d t

使用换元积分法

= \int \frac{1}{2 ( 1 + 81 u ^{2} )} 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 \frac{1}{1 + 81 u ^{2}} d u

使用换元积分法

= \frac{1}{2} \cdot \int \frac{1}{9 ( 1 + v ^{2} )} d v

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{9} \cdot \int \frac{1}{1 + v ^{2}} d v

使用积分常见定则:

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

= \frac{1}{2} \cdot \frac{1}{9} arctan (v)

代回

= \frac{1}{2} \cdot \frac{1}{9} arctan (9 t^{2})

化简

\frac{1}{2} \cdot \frac{1}{9} arctan (9 t^{2}) : \frac{1}{18} arctan (9 t^{2})

= \frac{1}{18} arctan (9 t^{2})

解答补常数

= \frac{1}{18} arctan (9 t^{2}) + C

作图

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