integral of (6t^3-t^2+24t)/(t^2+4)

解答

\int \frac{6 t ^{3} - t ^{2} + 24 t}{t ^{2} + 4} d t

3 t^{2} - t + 2 arctan (\frac{t}{2}) + 12 + C

求解步骤

\int \frac{6 t ^{3} - t ^{2} + 24 t}{t ^{2} + 4} d t

乘开

\frac{6 t ^{3} - t ^{2} + 24 t}{t ^{2} + 4} : \frac{6 t ^{3}}{t ^{2} + 4} - \frac{t ^{2}}{t ^{2} + 4} + \frac{24 t}{t ^{2} + 4}

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{6 t ^{3}}{t ^{2} + 4} d t - \int \frac{t ^{2}}{t ^{2} + 4} d t + \int \frac{24 t}{t ^{2} + 4} d t

\int \frac{6 t ^{3}}{t ^{2} + 4} d t = 3 (t^{2} + 4 - 4 ln t^{2} + 4)

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

\int \frac{24 t}{t ^{2} + 4} d t = 12 ln t^{2} + 4

= 3 (t^{2} + 4 - 4 ln t^{2} + 4) - (- 2 arctan (\frac{t}{2}) + t) + 12 ln t^{2} + 4

化简

3 (t^{2} + 4 - 4 ln t^{2} + 4) - (- 2 arctan (\frac{t}{2}) + t) + 12 ln t^{2} + 4 : 3 t^{2} - t + 2 arctan (\frac{t}{2}) + 12

= 3 t^{2} - t + 2 arctan (\frac{t}{2}) + 12

解答补常数

= 3 t^{2} - t + 2 arctan (\frac{t}{2}) + 12 + C

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