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anonymous
 5 years ago
How do I find the antiderivative of dx / (sqrt 14x^2) ?
anonymous
 5 years ago
How do I find the antiderivative of dx / (sqrt 14x^2) ?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use a substitution, 2x=sin(u), say.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then\[2dx=\cos u du \]and so\[\int\limits_{}{}\frac{dx}{\sqrt{14x^2}}=\frac{1}{2}\int\limits_{}{}\frac{\cos u}{\sqrt{1\sin^2 u}} du=\frac{1}{2}\int\limits_{}{} du=\frac{u}{2}+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then \[u=\sin^{1} \frac{x}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So the integral is\[\frac{1}{2}\sin^{1} \frac{x}{2}+ c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, \[\sin^{1} 2x\] I mean

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}{}\frac{dx}{\sqrt{14x^2}}=\frac{1}{2}\sin^{1} 2x +c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or the other easy way is to seprate deduction like:\[1\div \sqrt{14*x^2}=(A/\sqrt{12*x})+(B/\sqrt{1+2x})\] which A=1/sqrt{2}=B

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ijadi, your method's nice
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