integral of (t^5-1)/(t^4-1)

解答

\int \frac{t ^{5} - 1}{t ^{4} - 1} d t

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

求解步骤

\int \frac{t ^{5} - 1}{t ^{4} - 1} d t

乘开

\frac{t ^{5} - 1}{t ^{4} - 1} : \frac{t ^{5}}{t ^{4} - 1} - \frac{1}{t ^{4} - 1}

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 ^{5}}{t ^{4} - 1} d t - \int \frac{1}{t ^{4} - 1} d t

\int \frac{t ^{5}}{t ^{4} - 1} d t = \frac{t ^{2}}{2} - \frac{1}{4} (ln t^{2} + 1 - ln t^{2} - 1)

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

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

化简

\frac{t ^{2}}{2} - \frac{1}{4} (ln t^{2} + 1 - ln t^{2} - 1) - (- \frac{1}{2} arctan (t) - \frac{1}{4} ln ∣ t + 1 ∣ + \frac{1}{4} ln ∣ t - 1 ∣) : \frac{t ^{2}}{2} - \frac{1}{4} ln t^{2} + 1 + \frac{1}{4} ln t^{2} - 1 + \frac{1}{2} arctan (t) + \frac{1}{4} ln ∣ t + 1 ∣ - \frac{1}{4} ln ∣ t - 1 ∣

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

解答补常数

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

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