How do you integrate by substitution #int 1/sqrt(2x)dx#?

Answer 1

Ezekiel Alexander

Put #y=sqrt(2x)#
#x=y^2/2#
#dx=ydy#
#int1/sqrt(2x)dx=inty/ydy=y+c=sqrt(2x)+c#

You can do it without substitution remembering #intx^ndx=x^(n+1)/(n+1)+c;\ \ n=-1/2#

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Answer 2

Cameron Alexander

To integrate ( \int \frac{1}{\sqrt{2x}} , dx ), we use the substitution ( u = \sqrt{2x} ). This implies ( du = \frac{1}{\sqrt{2x}} , dx ). Substituting these into the integral gives:

( \int \frac{1}{\sqrt{2x}} , dx = \int du = u + C ),

where ( C ) is the constant of integration. Finally, substituting back ( u = \sqrt{2x} ) gives the result:

( \int \frac{1}{\sqrt{2x}} , dx = \sqrt{2x} + C ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.