How do you integrate #dx / (2sqrt(x) + 2x#?

1 Answer
Jim H
Mar 30, 2015

#int dx / (2sqrt(x) + 2x )#

Think about integration techniques, think about derivatives, think about fractions and factoring and . . .

There are "obvious" common factors in the denominator.

What do I know about #sqrtx# ? Among other things, I know the derivative is #1/(2sqrtx)#

I could factor out #2sqrtx# in the denominator. Will that help?
Try it and see. (Yes, I can try it in my head. Students will need to try it on paper.)

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

Did that help?

Not sure? write it as a product of 2 fracions where one fraction is the derivative of #sqrtx#

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

Now, what if I let #u=1+sqrtx#?

Then #du = 1/(2sqrtx) dx# and our integral becomes #int 1/u du# which is #lnu#, so we can quickly finish:

#int dx / (2sqrt(x) + 2x ) = ln(1+sqrtx) +C#

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