integral of (6t-3tx)/(t^2+1)

解答

\int \frac{6 t - 3 t x}{t ^{2} + 1}

3 ln 1 + t^{2} - \frac{3}{2} x ln 1 + t^{2} + C

求解步骤

\int \frac{6 t - 3 t x}{t ^{2} + 1} d t

乘开

\frac{6 t - 3 x t}{1 + t ^{2}} : \frac{6 t}{1 + t ^{2}} - \frac{3 x t}{1 + t ^{2}}

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

\int \frac{6 t}{1 + t ^{2}} d t = 3 ln 1 + t^{2}

\int \frac{3 x t}{1 + t ^{2}} d t = \frac{3}{2} x ln 1 + t^{2}

= 3 ln 1 + t^{2} - \frac{3}{2} x ln 1 + t^{2}

解答补常数

= 3 ln 1 + t^{2} - \frac{3}{2} x ln 1 + t^{2} + C

\int (1 - x^{3}) d x

\int 5 (400 + 5 t)^{3} d t

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\int 1 + 3 x^{3} (9 x^{2}) d x