How do you find the antiderivative of #int 1/root3(1-5t) dt#?

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Answer 1 · 2017-05-23T14:16:55

I got: #-3/10root3((1-5t)^2)+c#

Have a look:
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Answer 2 · 2017-05-24T12:59:54

#int(1/root3(1-5t))dt=-3/10(1-5t)^(2/3)+C#

#int(1/root3(1-5t))dt#

rewrite as

#int(1-5t)^(-1/3)dt#

now the outside of the bracket #=" constant "xx " bracket differentiated"#

so we are able to do this by inspection

using the power rule for integration ( add one to the power), and the fact that integration is the reverse of differentiation, let us try

#d/(dx)(1-5t)^(2/3)#

by the chain rule we get

#=2/3xx(-5)(1-5t)^(-1/3)=-10/3(1-5t)^(-1/3)#

so by comparing the integral and our 'inspected solution' we conclude

#int(1/root3(1-5t))dt=-3/10(1-5t)^(2/3)+C#