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The closet I can get is

$\arctan(x^4) $

the derivative of $\arctan(x^4)$ is $\dfrac{4x^3}{x^8 + 1} ...$

Any tips?

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up vote 10 down vote accepted

Caveat: The following explanation is for those who still have no knowledge about Integration/Anti-Derivative. This example emphasizes the fact of approaching an Integration with just the knowledge of differentiation

What you are trying to determine is called anti-derivative or Integration. In view of that consider the detailed explanation of how you should look over the problem. What we know $$\frac{df(x)}{dx}=\frac{4x^3}{1+x^4}\tag1$$ We need to determine what $f(x)$ is? Lets rewrite $(1)$ as $$df(x)=\frac{4x^3}{1+x^4}dx$$ Lets substitute $y = 1+x^4, dy = 4x^3dx$, then we have $$df(x)=\frac{dy}{y}\tag2$$ Now from your knowledge of derivative, you know that $$\frac{d}{dy}(\log y +C) = \frac{1}{y}$$ $$d(\log y +C) = \frac{dy}{y}\tag3$$ substituting $(3)\text{ on }(2)$ $$df(x)=d(\log(y) +C)$$ Substitute back $y = 1+x^4$ $$df(x)=d(\log(1+x^4)+C)$$ So $f(x)=\log(1+x^4)+C$

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Hint: Note that the derivative of $x^4 + 1\;$ is $4x^3$.

If we let $f(x) = x^4 + 1$, then we know $f'(x) = 4x^3$. Then note that $$\frac{4x^3}{x^4 + 1} = \dfrac{f'(x)}{f(x)}\tag{1}$$

And all integrals of the form $\displaystyle \int \dfrac{f'x}{f(x)}\,dx $ evaluate as $$\int \dfrac{f'x}{f(x)}\,dx = \ln|f(x)| + C \tag{where C is some constant}$$

  • Just note that $\;\frac{d}{dx}(\ln|f(x)| + C) = \dfrac{f'(x)}{f(x)}$ (by the chain rule).

If you already know how to integrate, you'll find we can integrate by substitution:

Let $u = f(x) =x^4 + 1$, then $f'(x) = du/dx = 4x^3 \implies \; du = 4x^3 \,dx$.

$$ \begin{align} \int \dfrac{4x^3\,dx}{x^4 + 1}\tag{By $(1)$} & = \int \dfrac{du}{u} = \ln|u| + C \\ \\ & = \ln(x^4 + 1) + C \\ \\ \end{align} $$

(because $x^4 + 1 \gt 0$, we don't need the absolute value sign: $|x^4 + 1|)$

If you haven't learned integration, note that $\; \dfrac{4x^3}{x^4 + 1}\;$ is precisely the derivative of $\ln(x^4 + 1)$.

You can check for yourself: using the chain rule, we get $$\frac d{dx}(\ln(x^4 + 1)) = \dfrac d{dx}(x^4 + 1)\cdot \dfrac {1}{x^4 + 1} = 4x^3 \cdot \dfrac{1}{x^4 + 1}$$

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koloa, does this make sense to you? –  amWhy Mar 19 '13 at 3:39
    
Thanks, @Babak: How are you? –  amWhy Mar 19 '13 at 14:15
    
Great! I see you are so fine today Amy.;-) –  B. S. Mar 19 '13 at 14:24
    
Just woke up an hour ago :-/ $\quad$ I was up very late last night, hoping to see you in your morning, but alas, no Babak! So good to see you now! –  amWhy Mar 19 '13 at 14:30
    
Thanks for your kind words. ;-) –  B. S. Mar 19 '13 at 14:47
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HINT: Just substitute $u=x^4+1$.

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Basically you want to find

$$\int dx \: \frac{4 x^3}{x^4+1}$$

Note that

$$d(x^4) = 4 x^3 dx$$

so that the integral is

$$\int \frac{d(x^4)}{x^4+1} = \log{(x^4+1)} + C$$

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Would it be ok to write $d(x^4 +1) = 4x^3$ in the "Note"? –  The Chaz 2.0 Mar 19 '13 at 4:12
    
@TheChaz2.0 Apart from the missing $dx$ yes, since constants don't vary –  Tobias Kienzler Mar 19 '13 at 8:03
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