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I need some suggestion how to solve this integral.
$$\int^1_0 \frac{xdx}{x^2+2x+1}$$ I think about to do the following step :
$$\frac{1}{2}\int^1_0\frac{2x+2-2dx}{x^2+2x+1}$$$$ t=x^2+2x+1 \rightarrow 2x+2dx=dt$$ then the integral will be : $$\frac{1}{2}\int^1_0 \frac{-2dt}{t}$$ its a correct way to solve it?
Thanks!

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Hint:

Notice that $x^2+2x+1=(x+1)^2$

$\int^1_0 \dfrac{xdx}{x^2+2x+1}=\int_0^1\dfrac{A}{x+1}+\int_0^1\dfrac{B}{(x+1)^2}$

What is wrong with your approach?

After you have $dt=2(x+1)dx$, your expression will be $\dfrac{1}{2} \int_0^1 \dfrac{2(\sqrt{t}-1)}{2\sqrt{t} \cdot t} dt$

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  • $\begingroup$ I get : $\int_0^1\dfrac{1}{x+1}+\int_0^1\dfrac{-1}{(x+1)^2}$ right? $\endgroup$ – Ofir Attia May 14 '13 at 8:23
  • $\begingroup$ @OfirAttia: Yes. That's right. $\endgroup$ – Inceptio May 14 '13 at 8:26
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@Ofir, you are almost right:

$$\frac{1}{2}\int^1_0\frac{2x+2-2dx}{x^2+2x+1}=\frac{1}{2}\left(\int^1_0\frac{2x+2}{x^2+2x+1}dx+\int^1_0\frac{-2}{x^2+2x+1}dx\right)=\frac{1}{2}(\ln|{x^2+2x+1}|_{0}^{1}+\frac{2}{1+x}|^{1}_{0})=\frac{1}{2}(\ln{4}-\ln{1}+1-2)=\frac{1}{2}(\ln^4-1)$$

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You followed the general strategy. Alternatively, you can remark that $x^2+2x+1=(x+1)^2$,, put $u=x+1$ and reduce to $$\int \frac{u-1}{u^2}\, du,$$ which is really simple.

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  • $\begingroup$ OK, but just for my aknowledge what I did its good or its wrong way to do that? $\endgroup$ – Ofir Attia May 14 '13 at 7:52
  • $\begingroup$ The method is standard and legitimate: you did some algebraic tricks to get the derivative of the denominator at the numerator. If your answer is actually a primitive, it's ok. $\endgroup$ – Siminore May 14 '13 at 7:55
  • $\begingroup$ yes, but he did a few mistakes! $\endgroup$ – Wolphram jonny May 14 '13 at 8:17
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$$\int^1_0 \frac{xdx}{x^2+2x+1}$$ $$\int^1_0 \frac{x}{(x+1)^2}\, dx$$ $$\int^1_0 \frac{x+1-1}{(x+1)^2}\, dx$$ $$\int^1_0 \frac{1}{(x+1)}-\frac{1}{(x+1)^2}\, dx$$ $$[\log(x+1)+\frac{1}{x+1}]_0^1$$ $$\log 2-\frac{1}{2}$$

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You started well, but ended it wrong. The numerator is $(2x+2)dx-2dx=dt-2dx$, in the second term I left $dx$ because the change of variables is not useful to it, so you are left with $\frac{1}{2}\int^4_0\frac{dt}{t}-\int^1_0\frac{dx}{x^2+2x+1}$. Notice also that you also forgot to change the limits of the first integral, that is now over $t$, that is why it ranges from 0 to 4 instead than from 0 to 1. The second integral is easy, but let me know if you cannot solve it (hint: is equal to $\int^2_1\frac{dt}{t^2}$)

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  • $\begingroup$ the second integral is the pattern of arctanx? $\endgroup$ – Ofir Attia May 14 '13 at 8:08
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    $\begingroup$ no, for arctan you have to have $1/(1+x^2)$, but the one you have is the same as $1/(x+1)^2$, which is different, but you can use a change of variables $t=x+1$ $\endgroup$ – Wolphram jonny May 14 '13 at 8:11

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