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integration 1/x-?x
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Asked by dubeynitesh9876 | 28 Feb, 2024, 09:51: AM
Hence f(x) = g(x) + h (x)
where g(x) = x / [ x2 - x ] = 1/(x-1) and h(x) = √x / [ x2 - x ] = 1 / [ √x (x-1) ]
Let u = √x
du = [ 1 / ( 2 √x ) ] dx
Hence ( dx / √x ) = 2 du
Hence , we rewrite the second integration as
If we write the second integrand function as partial fracion , then we have
Hence above integration becomes
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