结果1 结果2 结果3 结果4 结果5 题目运用第二类换元积分法计算下列不定积分:$$ \int \frac { 1 } { x \sqrt { x ^ { 2 } - 1 } } d x ; $$ 相关知识点: 试题来源: 解析 $$ \arccos \frac { 1 } { | x | } + C $$
【解析】 方法一:当$$ x > 1 $$时,令$$ x = s e c t , t \in ( 0 , \frac { \pi } { 2 } ) $$,则$$ \frac { 1 } { x } = \cos t , d x = s e c t \tan t d t , $$ 于是 $$ \int \frac { 1 } { x \sqrt { x ^ { 2 } - 1 } }...
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Answer to: integrate \int \frac{1}{x^{3}\sqrt{x^{2}-1}}dx,x> 1 By signing up, you'll get thousands of step-by-step solutions to your homework...
1判定下列各反常积分的收敛性,如果收敛,计算反常积分的值: 2判定下列各反常积分的收敛性,如果收敛,计算反常积分的值:$$ \int _ { 0 } ^ { 1 } \frac { d x } { ( 2 - x ) \sqrt { 1 - x } } $$ 3判定下列各反常积分的收敛性,如果收敛,计算反常积分的值∫_1^(+∞)(ln^2x...
解析 \$\int \frac { 1 } { \sqrt { x } } \mathrm { d } x = \int x ^ { - \frac { 1 } { 2 } } \mathrm { d } x = \frac { x ^ { - \frac { 1 } { 2 } + 1 } } { - \frac { 1 } { 2 } + 1 } + C = 2 \sqrt { x } + C\$ ...
解: $$ 设t= \sqrt{2x} \\ x= \frac{t^{2}}{2} $$ $$ \frac{1}{\sqrt{2x}+1}dx \\ = \left\{ \frac{1}{t+1}d \frac{t^{2}}{2} \\ =| \frac{t}{t+1}dt \\ = \int(1- \frac{1}{t+1})dt $$ $$ =t-In(t+1)= \sqrt{2x}-In(\sqrt{2}x...
Answer to: Integrate : \int \frac{dx}{\sqrt{x^{2}-1}} By signing up, you'll get thousands of step-by-step solutions to your homework questions. You...
$$ 1 、 \int \frac { a x } { 1 + \sqrt { 2 x } } $$ 答案 解$$ \int \frac{dx}{1+ \sqrt{2x}} $$ 令$$ t= \sqrt{2x} $$即. $$ x= \frac{1}{2}t^{2} $$ $$ d_{x}=d(\frac{1}{2}t^{2})=tdt $$ $$原式 原式= \int \frac{t}{1...
【解析】 解 因的$$ \sqrt { 1 + x ^ { 2 } } $$的是复合函数,设$$ u ( = \varphi ( x ) ) = 1 + x ^ { 2 } , ( 1 + x ^ { 2 } ) ^ { \prime } d x = 2 x d x = $$ $$ d ( 1 + x ^ { 2 } $$.故 $$ \int \frac { x } { \sqrt...