integral of (x^2)/(1+25x^2)

解答

\int \frac{x ^{2}}{1 + 25 x ^{2}} d x

\frac{1}{125} (- arctan (5 x) + 5 x) + C

求解步骤

\int \frac{x ^{2}}{1 + 25 x ^{2}} d x

使用换元积分法

= \int \frac{u ^{2}}{125 ( 1 + u ^{2} )} d u

提出常数:

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

= \frac{1}{125} \cdot \int \frac{u ^{2}}{1 + u ^{2}} d u

\frac{u ^{2}}{1 + u ^{2}} = - \frac{1}{1 + u ^{2}} + 1

= \frac{1}{125} \cdot \int - \frac{1}{1 + u ^{2}} + 1 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

= \frac{1}{125} (- \int \frac{1}{1 + u ^{2}} d u + \int 1 d u)

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

\int 1 d u = u

= \frac{1}{125} (- arctan (u) + u)

u = 5 x

代回

= \frac{1}{125} (- arctan (5 x) + 5 x)

解答补常数

= \frac{1}{125} (- arctan (5 x) + 5 x) + C

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