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I'm trying to integrate $$\frac{8x^2+3x+1}{x(2x+1)^2}$$

I did a partial fraction expansion: $$\frac{8x^2+3x+1}{x(2x+1)^2}= \frac{1}{x}+\frac{2}{2x+1}-\frac{3}{(2x+1)^2}$$

and now I'm left with $$\int\left(\frac{1}{x}+\frac{2}{2x+1}-\frac{3}{(2x+1)^2}\right)dx$$

I would get $\ln(x) + \ln(2x+1)$ from integrating $\frac{1}{x}+\frac{2}{2x+1}$ but I do not quite know how to integrate $$ \frac{3}{(2x+1)^2} $$

It would be great if someone can teach me what steps I need to take to integrate $\frac{3}{(2x+1)^2}$, thanks.

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    $\begingroup$ If you choose $u = 2x + 1$, what is $du$? How do it compare with the '$3 dx$' you have in the numerator of the integrand? How can you now write the integral? $\endgroup$ – colormegone Apr 20 '13 at 4:46
  • $\begingroup$ @RecklessReckoner Thanks! I get it now. $\endgroup$ – 吖奇说 ARCHY SHUō Apr 20 '13 at 5:25
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We want to integrate an integral of the form $$\int \dfrac{dx}{(ax+b)^n}$$ Let $y=ax+b$. We then have $dy = a dx \implies dx = \dfrac{dy}a$ $$\int \dfrac{dx}{(ax+b)^n} = \int \dfrac{dy}{ay^n} = \dfrac{y^{-n+1}}{a(-n+1)} + \text{constant} = \dfrac1{a(1-n)(ax+b)^{n-1}} + \text{constant}$$

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Substitute $u=2x+1 \implies du=2dx $. Then the integral is $$\int\frac{3}{2}\frac{1}{u^2}du=\frac{-3}{2u}+C'=\frac{-3}{4x+2}+C.$$

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