integral of ((6t-8)/((t-3)^3))

解答

\int (\frac{6 t - 8}{( t - 3 ) ^{3}}) d t

- \frac{6}{t - 3} - \frac{5}{( t - 3 ) ^{2}} + C

求解步骤

\int \frac{6 t - 8}{( t - 3 ) ^{3}} d t

乘开

\frac{6 t - 8}{( t - 3 ) ^{3}} : \frac{6 t}{( t - 3 ) ^{3}} - \frac{8}{( t - 3 ) ^{3}}

= \int \frac{6 t}{( t - 3 ) ^{3}} - \frac{8}{( t - 3 ) ^{3}} 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{6 t}{( t - 3 ) ^{3}} d t - \int \frac{8}{( t - 3 ) ^{3}} d t

\int \frac{6 t}{( t - 3 ) ^{3}} d t = 6 (- \frac{1}{t - 3} - \frac{3}{2 ( t - 3 ) ^{2}})

\int \frac{8}{( t - 3 ) ^{3}} d t = - \frac{4}{( t - 3 ) ^{2}}

= 6 (- \frac{1}{t - 3} - \frac{3}{2 ( t - 3 ) ^{2}}) - (- \frac{4}{( t - 3 ) ^{2}})

化简

6 (- \frac{1}{t - 3} - \frac{3}{2 ( t - 3 ) ^{2}}) - (- \frac{4}{( t - 3 ) ^{2}}) : - \frac{6}{t - 3} - \frac{5}{( t - 3 ) ^{2}}

= - \frac{6}{t - 3} - \frac{5}{( t - 3 ) ^{2}}

解答补常数

= - \frac{6}{t - 3} - \frac{5}{( t - 3 ) ^{2}} + C

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