- #1

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ok so here is the problem

[tex]\int\frac{dx}{\sqrt(x^2-4x)}[/tex]

the table integral I am supposed to use is this

[tex]\int\frac{du}{\sqrt{u^2-a^2}}=ln(u+\sqrt{u^2-a^2}+C[/tex]

Is it proper to make my u=x^2 and a=2x^(1/2)

I am asking because the solution guide tells me to complete the square, and then proceed to pick the u and a

[tex]\int\frac{dx}{\sqrt(x^2-4x)}[/tex]

the table integral I am supposed to use is this

[tex]\int\frac{du}{\sqrt{u^2-a^2}}=ln(u+\sqrt{u^2-a^2}+C[/tex]

Is it proper to make my u=x^2 and a=2x^(1/2)

I am asking because the solution guide tells me to complete the square, and then proceed to pick the u and a

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