integral of (8t^3+10)/(4t^2-4t+5)

解答

\int \frac{8 t ^{3} + 10}{4 t ^{2} - 4 t + 5} d t

t^{2} + 2 t - \frac{1}{4} ln 4 t^{2} - 4 t + 5 - \frac{1}{4} arctan (t - \frac{1}{2}) + C

求解步骤

\int \frac{8 t ^{3} + 10}{4 t ^{2} - 4 t + 5} d t

长除

\frac{8 t ^{3} + 10}{4 t ^{2} - 4 t + 5} : 2 t + 2 - \frac{2 t}{4 t ^{2} - 4 t + 5}

= \int 2 t + 2 - \frac{2 t}{4 t ^{2} - 4 t + 5} 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 2 t d t + \int 2 d t - \int \frac{2 t}{4 t ^{2} - 4 t + 5} d t

\int 2 t d t = t^{2}

\int 2 d t = 2 t

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

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

化简

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

解答补常数

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

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