What Is the Correct Substitution for the Integral of 4x/(1+4x^2)?

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How do you get that the integral of 4x/(1+4x^2) = 1/2*ln(1_4x^2) ?
I have no clue. I tried u substitution with y=4x^2 but that got me nowhere.
Thanks!
 
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Note that:

[tex] \frac{4x}{1+4x^{2}}=\frac{1}{2}\frac{8x}{1+4x^{2}}=\frac{1}{2}\frac{f'(x)}{f(x)}[/tex]

Where [tex]f(x)=1+4x^{2}[/tex]
 
sparkle123 said:
How do you get that the integral of 4x/(1+4x^2) = 1/2*ln(1_4x^2) ?
I have no clue. I tried u substitution with y=4x^2 but that got me nowhere.
Thanks!

Even though you could have included the 1 in the substitution, the one you chose would have solved your problem the same way since

[tex]\int \frac{du}{1+u} = \ln (1+u) + C[/tex] , whenever u>0.