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I'm learning integrals and I have one question.

I have this integral:

$$\int\frac{2x}{1+x^2}$$

I thought the solution was using $f'/(1+f^2)$ letting $f'=2x$ and $f^2 = x^2$ so the solution would be $\arcsin(x) + C$.

Searching for the right answer on the internet, everyone uses $$\frac{f'}{f}$$ so the solution is $\log|1+x^2| + C$

Can someone explain me why my solution is wrong?

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3 Answers

up vote 2 down vote accepted

The first solution is wrong while the second is correct.

You have $f(x)=1+x^2$ when you "let $f^{\prime}(x)=2x$" (*) not $f^2(x)=x^2$!! Thus it is $$\frac{f^{\prime}}{f}$$ and not $$\frac{f^{\prime}}{1+f^2}$$ If the second were true, then indeed a primitive would have been $\arcsin f$

(*) Usually, you don't actually let $f^{\prime}=...$ but rather $f=...$ and then you compute $f^{\prime}$

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Hi Nameless. Thanks for your information. Now I understand. –  Favolas Dec 29 '12 at 12:24
    
@Favolas Glad to help –  Nameless Dec 29 '12 at 12:25
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$f^2\neq f.f$ Instead your $f$ should be $x^2.$ so that you can get $f'(x)=2x$

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$$\frac{d\arcsin x}{dx}=\frac1{\sqrt{1-x^2}}$$

So, $$\int\frac{dx}{\sqrt{1-x^2}}=\arcsin x+C$$, so the first solution is wrong.

$$\int \frac{f'}{1+f^2}dx=\int \frac{df}{1+f^2}=\arctan f+C$$

and $$\int \frac{f'}fdx=\int \frac{df}f=\log f+C$$ as $f'dx=\frac{df}{dx}dx=df$

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