integral from 0 to x of (t^2)/(t^4+1)

解答

\int_{0}^{x} \frac{t ^{2}}{t ^{4} + 1} d t

- \frac{1}{4 2} ln 2 x^{2} + 22 x + 2 + \frac{1}{4 2} ln 2 x^{2} - 22 x + 2 + \frac{1}{2 2} arctan (2 x + 1) + \frac{1}{2 2} arctan (2 x - 1)

求解步骤

\int_{0}^{x} \frac{t ^{2}}{t ^{4} + 1} d t

将

\frac{t ^{2}}{t ^{4} + 1}

用部份分式展开:

- \frac{t}{2 2 ( t ^{2} + 2 t + 1 )} + \frac{t}{2 2 ( t ^{2} - 2 t + 1 )}

= \int_{0}^{x} - \frac{t}{2 2 ( t ^{2} + 2 t + 1 )} + \frac{t}{2 2 ( t ^{2} - 2 t + 1 )} d t

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

= - \int_{0}^{x} \frac{t}{2 2 ( t ^{2} + 2 t + 1 )} d t + \int_{0}^{x} \frac{t}{2 2 ( t ^{2} - 2 t + 1 )} d t

\int_{0}^{x} \frac{t}{2 2 ( t ^{2} + 2 t + 1 )} d t = \frac{1}{2} (\frac{1}{2 2} (ln 2 x^{2} + 22 x + 2 - ln (2)) - \frac{1}{2} (arctan (2 x + 1) - \frac{π}{4}))

\int_{0}^{x} \frac{t}{2 2 ( t ^{2} - 2 t + 1 )} d t = \frac{1}{2} (\frac{1}{2 2} (ln 2 x^{2} - 22 x + 2 - ln (2)) + \frac{1}{2} (arctan (2 x - 1) + \frac{π}{4}))

= - \frac{1}{2} (\frac{1}{2 2} (ln 2 x^{2} + 22 x + 2 - ln (2)) - \frac{1}{2} (arctan (2 x + 1) - \frac{π}{4})) + \frac{1}{2} (\frac{1}{2 2} (ln 2 x^{2} - 22 x + 2 - ln (2)) + \frac{1}{2} (arctan (2 x - 1) + \frac{π}{4}))

化简

- \frac{1}{2} (\frac{1}{2 2} (ln 2 x^{2} + 22 x + 2 - ln (2)) - \frac{1}{2} (arctan (2 x + 1) - \frac{π}{4})) + \frac{1}{2} (\frac{1}{2 2} (ln 2 x^{2} - 22 x + 2 - ln (2)) + \frac{1}{2} (arctan (2 x - 1) + \frac{π}{4})) : - \frac{1}{4 2} ln 2 x^{2} + 22 x + 2 + \frac{1}{4 2} ln 2 x^{2} - 22 x + 2 + \frac{1}{2 2} arctan (2 x + 1) + \frac{1}{2 2} arctan (2 x - 1)

= - \frac{1}{4 2} ln 2 x^{2} + 22 x + 2 + \frac{1}{4 2} ln 2 x^{2} - 22 x + 2 + \frac{1}{2 2} arctan (2 x + 1) + \frac{1}{2 2} arctan (2 x - 1)

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