How do you evaluate #int(1-x)/(1-sqrtx)dx#?

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Answer 1 · 2018-07-18T14:42:47

#x+2/3x^{3/2}+C#

#\int \frac{1-x}{1-\sqrtx}\ dx#

#=\int \frac{1^2-(\sqrtx)^2}{1-\sqrt{x}}\ dx#

#=\int \frac{(1+\sqrtx)(1-\sqrtx)}{1-\sqrt{x}}\ dx#

#=\int (1+\sqrtx)\ dx#

#=\int 1\ dx+\int \sqrtx\ dx#

#=x+\int x^{1/2}\ dx#

#=x+\frac{x^{1/2+1}}{1/2+1}+C#

#=x+2/3x^{3/2}+C#