integral of (t^2)/(t^4+1)

解答

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

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

求解步骤

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

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

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

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

解答补常数

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

\frac{\partial}{\partial y} (6 e^{yz} cos (x yz))

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