integral of 1/(t+t^3)

解答

\int \frac{1}{t + t ^{3}} d t

ln ∣ t ∣ - \frac{1}{2} ln t^{2} + 1 + C

求解步骤

\int \frac{1}{t + t ^{3}} d t

将

\frac{1}{t + t ^{3}}

用部份分式展开:

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

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

\int \frac{1}{t} d t = ln ∣ t ∣

\int \frac{t}{t ^{2} + 1} d t = \frac{1}{2} ln t^{2} + 1

= ln ∣ t ∣ - \frac{1}{2} ln t^{2} + 1

解答补常数

= ln ∣ t ∣ - \frac{1}{2} ln t^{2} + 1 + C

\int ln (e^{2 t - 1}) d t

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