按部件集成 - 是否有更简单的方法来设置u和dv?

Whistlekins 06/21/2018. 3 answers, 61 views
integration

有问题的积分是$$ \ int(2-x)^ 2 \ ln {(4x)} dx $$

我用$ du = \ frac {4} {x} dx $和$ dv =(2-x)^ 2dx $设置$ u = \ ln {(4x)} $,$ v = - \ frac {1 } {3}(2-x)^ 3 $

使用部件集成,我最终得到:

$$ - \压裂{1} {3}(2-x)^ 3 \ {LN(4倍)} - \ INT \压裂{-4} {3×}(2-x)^ 3DX $$

$$ = - \压裂{1} {3}(2-x)^ 3 \ {LN(4×)} + \压裂{32} {3} \ {LN(X)} - 16X + 4X ^ 2- \压裂{4} {9}的x ^ 3 + C $$

我不知道我现在是否正确(没有任何答案),但我只是想知道将$ u $和$ dv $设置为其他东西会更容易。 我已经尝试交换原来的$ u $和$ dv $,但这似乎更难。

请记住,这是一个按部件问题的集成,所以使用另一种方法并不是我真正想要的。

编辑:谢谢大家,我现在看到了我的错误。

3 Answers


farruhota 06/21/2018.

错误在于:$$ \ left(\ ln(4x)\ right)'= \ frac {1} {4x} \ cdot(4x)'= \ frac {1} {4x} \ cdot 4 = \ frac1x。 $$所以通过你的方法:$$ \ int(2-x)^ 2 \ ln {(4x)} dx = \ int \ underbrace {\ ln {(4x)}} _ {u} \ cdot \ underbrace {( 2-x)^ 2dx} _ {dv} = \\ = \ underbrace {\ ln(4x)} _ {u} \ cdot \ underbrace {\ frac { - (2-x)^ 3} {3}} _ {v} - \ int \ underbrace {\ frac { - (2-x)^ 3} {3}} _ {v} \ cdot \ underbrace {\ frac1xdx} _ {du} = \\ - \ frac {(2 -x)^ 3} {3} \ cdot \ ln(4x)+ \ frac {8} {3} \ ln {(x)} - 4x + x ^ 2- \ frac {1} {9} x ^ 3 + C. $$可以简化如下:$$ I = \ require {cancel} - \ frac {8} {3} \ cdot \ ln(4x)+ \ left(4x-2x ^ 2 + \ frac { x ^ 3} {3} \ right)\ cdot \ ln(4x)+ \ frac {8} {3} \ ln {(x)} - 4x + x ^ 2- \ frac {1} {9} x ^ 3 + C = \\ \ color {red} { - \ frac {8} {3} \ cdot \ ln(4)} \ cancel { - \ frac {8} {3} \ cdot \ ln(x)} + \ left(4x-2x ^ 2 + \ frac {x ^ 3} {3} \ right)\ cdot \ ln(4x)\ cancel {+ \ frac {8} {3} \ ln {(x)}} - 4x + x ^ 2- \ frac {1} {9} x ^ 3 + \ color {red} C = \\ \ left(4x-2x ^ 2 + \ frac {x ^ 3} {3} \ right)\ cdot \ ln(4x)-4x + x ^ 2- \ frac {1} {9} x ^ 3 + C. $$注意:如果使用它会更短:$$ dv =(2-x)^ 2dx \ Rightarrow dv =(4-4x + x ^ 2)dx \ Rightarrow v = 4x-2x ^ 2 + \ frac {x ^ 3} {3}。$$


$$ \ INT(2-x)^ 2 \ {LN(4×)} DX $$

按部件集成

$$ f = \ ln {(4x)},\ quad g'=(x-2)^ 2 $$ $$ f'= \ frac {1} {x},\ quad g = \ frac {x ^ 3 } {3} -2x ^ 2 + 4x $$这跟随$$ \ left(\ frac {x ^ 3} {3} -2x ^ 2 + 4x \ right)\ ln(4x) - \ underbrace {\ int \ frac {\ frac {x ^ 3} {3} -2x ^ 2 + 4x} {x} \,dx} _ {I} $$

\ begin {align} I&= \ int \ left(\ frac {x ^ 2} {3} -2x + 4 \ right)dx \\&= \ int \ frac {x ^ 2} {3} \,dx -2 \ int x \,dx + \ int4 \,dx \\&= \ frac {x ^ 3} {9} -x ^ 2 + 4x \ end {align}

所以

$$ \ left(\ frac {x ^ 3} {3} -2x ^ 2 + 4x \ right)\ ln(4x) - \ frac {x ^ 3} {9} + x ^ 2-4x + C $$


user570193 06/21/2018.

不要替代。

$ \ int(4-4x + x ^ 2)\ ln(4x)dx =(4x-2x ^ 2 + \ frac {1} {3} x ^ 3)\ ln(4x) - \ int \ frac {( 4X-2X ^ 2 + \压裂{1} {3}的x ^ 3)} {X} DX $

$ =(4x-2x ^ 2 + \ frac {1} {3} x ^ 3)\ ln(4x) - \ int(4-2x + \ frac {1} {3} x ^ 2)dx $

你可以从这里拿走它。


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