# Variable Change in Basic Integration

Could someone please explain to me how to do basic integration using the change variable method?

I understand how to find "u" in the integrand, and then find du/dx, but I don't really know what's going on after that.

For example:

$$\int$$ $$\sqrt{ln(x)}$$/x

u= ln(x), right?
and du/dx = 1/x

I know that the integral is equal to $$\sqrt{u}$$= (2/3) ln(x)^(3/2)
But WHY? What happens to 1/x? Is it ignored becuase it's the derivative of ln(x)?

Thanks

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## Answers and Replies

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rl.bhat
Homework Helper
Could someone please explain to me how to do basic integration using the change variable method?

I understand how to find "u" in the integrand, and then find du/dx, but I don't really know what's going on after that.

For example:

$$\int$$ $$\sqrt{ln(x)}$$/x

u= ln(x), right?
and du/dx = 1/x

I know that the integral is equal to $$\sqrt{u}$$= (2/3) ln(x)^(3/2)
But WHY? What happens to 1/x? Is it ignored becuase it's the derivative of ln(x)?

Thanks
$$\int$$ $$\sqrt{ln(x)}$$/x
It should be written as
$$\int$$ $$\sqrt{ln(x)}$$/x*dx
lnx = u
du = 1/x*dx