integral of (2t^3-t^2+2t+4)/(1+t^2)

解答

\int \frac{2 t ^{3} - t ^{2} + 2 t + 4}{1 + t ^{2}} d t

t^{2} - t + 5 arctan (t) + C

求解步骤

\int \frac{2 t ^{3} - t ^{2} + 2 t + 4}{1 + t ^{2}} d t

长除

\frac{2 t ^{3} - t ^{2} + 2 t + 4}{1 + t ^{2}} : 2 t - 1 + \frac{5}{t ^{2} + 1}

= \int 2 t - 1 + \frac{5}{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 2 t d t - \int 1 d t + \int \frac{5}{t ^{2} + 1} d t

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

\int 1 d t = t

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

= t^{2} - t + 5 arctan (t)

解答补常数

= t^{2} - t + 5 arctan (t) + C