How do you integrate $\frac{d x}{2 \sqrt{x} + 2 x}$ $dx / (2sqrt(x) + 2x$ ?

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How do you integrate $\frac{d x}{2 \sqrt{x} + 2 x}$ $dx / (2sqrt(x) + 2x$ ?

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Answer 1 · 2015-03-30T16:16:08

$\int \frac{d x}{2 \sqrt{x} + 2 x}$ $int dx / (2sqrt(x) + 2x )$

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 $\sqrt{x}$ $sqrtx$ ? Among other things, I know the derivative is $\frac{1}{2 \sqrt{x}}$ $1/(2sqrtx)$

I could factor out $2 \sqrt{x}$ $2sqrtx$ 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.)

$\int \frac{d x}{2 \sqrt{x} + 2 x} = \int \frac{1}{2 \sqrt{x} (1 + \sqrt{x})} d x$ $int dx / (2sqrt(x) + 2x ) = int 1/(2sqrtx(1+sqrtx))dx$

Did that help?

Not sure? write it as a product of 2 fracions where one fraction is the derivative of $\sqrt{x}$ $sqrtx$

$\int \frac{d x}{2 \sqrt{x} + 2 x} = \int \frac{1}{2 \sqrt{x}} \frac{1}{1 + \sqrt{x}} d x$ $int dx / (2sqrt(x) + 2x ) = int 1/(2sqrtx) 1/(1+sqrtx)dx$

Now, what if I let $u = 1 + \sqrt{x}$ $u=1+sqrtx$ ?

Then $d u = \frac{1}{2 \sqrt{x}} d x$ $du = 1/(2sqrtx) dx$ and our integral becomes $\int \frac{1}{u} d u$ $int 1/u du$ which is $ln u$ $lnu$ , so we can quickly finish:

$\int \frac{d x}{2 \sqrt{x} + 2 x} = ln (1 + \sqrt{x}) + C$ $int dx / (2sqrt(x) + 2x ) = ln(1+sqrtx) +C$

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How do you integrate $\frac{d x}{2 \sqrt{x} + 2 x}$ $dx / (2sqrt(x) + 2x$ ?

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