integral from 0 to t of 6/(1+6t^2)

解答

\int_{0}^{t} \frac{6}{1 + 6 t ^{2}} d t

6 arctan (6 t)

求解步骤

\int_{0}^{t} \frac{6}{1 + 6 t ^{2}} d t

提出常数:

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

= 6 \cdot \int_{0}^{t} \frac{1}{1 + 6 t ^{2}} d t

使用换元积分法

= 6 \cdot \int_{0}^{6 t} \frac{1}{6 ( 1 + u ^{2} )} d u

提出常数:

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

= 6 \cdot \frac{1}{6} \cdot \int_{0}^{6 t} \frac{1}{1 + u ^{2}} d u

使用积分常见定则:

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

= 6 \cdot \frac{1}{6} [arctan (u)]_{0}^{6 t}

化简

6 \cdot \frac{1}{6} [arctan (u)]_{0}^{6 t} : 6 [arctan (u)]_{0}^{6 t}

= 6 [arctan (u)]_{0}^{6 t}

带入上下限计算后得:

arctan (6 t)

= 6 arctan (6 t)

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