How do you integrate #int (1) / (sqrt(1 + x))#?

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Answer 1 · 2018-05-28T14:27:37

#int1/sqrt(x+1)dx=2sqrt(x+1)+c#

#int1/sqrt(x+1)dx=2int((x+1)')/(2sqrt(x+1))dx=#

#2int(sqrt(x+1))'dx=2sqrt(x+1)+c# #color(white)(aa)# , #c##in##RR#

Answer 2 · 2018-05-28T14:28:15

#2sqrt(1+x)+C#

This function is very close to #sqrt(\frac{1}{x})#, whose integral is #2sqrt(x)#. In fact,

#\frac{d}{dx} 2sqrt(x) = 2\frac{d}{dx} sqrt(x) =2\frac{1}{2sqrt(x)} = \frac{1}{sqrt(x)}#

In our integral, you can substitute #t=x+1#, which implies #dt=dx#, since this is only a translation. So, you'd have

#\int \frac{1}{sqrt(t)} dt = 2sqrt(t)+C = 2sqrt(1+x) + C#