integral of dx/(1-x*sqrt(x))
\[\int\limits_{}^{} \frac{ dx }{ 1-x \sqrt{x} }\]
try \(\sqrt{x} = u \)
\(\implies x = u^2 \implies dx = 2u~du\)
\[\large \int \dfrac{2u~du}{1-u^3}\]
partial fractions etc..
\[\int\limits_{}^{}\frac{ 1 }{ 1-u^2 } \] equals to \[\frac{ 1 }{2 } \ln \frac{ 1+u }{ 1-u }\] but what happends to \[\int\limits_{}^{}2u\] because the answer in my books is \[-\ln \frac{ \sqrt{x}-1 }{ \sqrt{x}+1 } +c\]
if possible, can you take a screenshot and attach..
can't...writing from computer, and dont have cable for my phone to connect pics with. but there is a page with identities of diffrent laws that states like this: \[\frac{ 1 }{ a^2-x^2 }\] where a^2 = 1^2 and x^2 = u^2 and that function has an indefinite integral of \[\frac{ 1 }{ 2a } \ln \frac{ a+x }{ a-x } +c\] a =1 and x = sqrt(x)
that identity looks correct, but it may not be useful for our present problem
wolfram gives this : http://www.wolframalpha.com/input/?i=%5Cint+1%2F%281-xsqrt%28x%29%29+dx
thanx
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