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Completely stuck. What is the easiest way to approach:

$$ \int \frac{1}{\left(x^{2}+7 x+4\right)^{2}} d x $$

I tried substitution, partial fractions, etc.

Update:
When I try partial fractions,

$$ \frac{A x+B}{x^{2}+7 x+4}+\frac{C x+D}{\left(x^{2}+7 x+4\right)^{2}}$$

I end up back at the starting point.

$$ \frac{0x+0}{x^{2}+7 x+4}+\frac{0 x+1}{\left(x^{2}+7 x+4\right)^{2}}$$

I can't see what I am doing wrong with partial fractions.

user41592
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  • Factor $x^2+7x+4$, then use partial fraction. – Just a user Feb 20 '22 at 10:59
  • How do I factor $x^2+7x+4$ for partial fractions? – user41592 Feb 20 '22 at 11:04
  • It's quite annoying to get downvotes without any justification (in my answer) clearly my answer contains a lot of pending task for the OP. – A. P. Feb 20 '22 at 12:03
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    "Partial fractions" is exactly what works, so you were in the right direction, although I don't know how far you went or where you went wrong. It is likely, I think , that you may not have realised that the "factorization" of $(x^2+7x+4)^2$ isn't a product of integer-coeffficient polynomials , but rather of real-coefficient polynomials. It is possible to provide an abstract expression for the integral of rational functions in terms of roots, see here. – Sarvesh Ravichandran Iyer Feb 20 '22 at 12:11

1 Answers1

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We have a polynomial in the integran so we should try complete the square and then partial fraction.

\begin{align*} \int \frac{1}{(x^{2}+7x+4)^{2}}{\rm d}x&=\int \frac{1}{((x+7/2)^{2}-33/4)^{2}}{\rm d}x\\&\overset{x\mapsto x+7/2}{=}\int \frac{1}{(x^{2}-33/4)^{2}}{\rm d}x\\&=\int \left(\frac{4}{33\sqrt{33}(2x+\sqrt{33})}+\frac{4}{33(2x+\sqrt{33})^{2}}+\frac{4}{33\sqrt{33}(-2x+\sqrt{33})}+\frac{4}{33(-2x+\sqrt{33})^{2}}\right){\rm d}x \end{align*}

The last primitive is just using change of variables, solving and substitute back we have $$ \int \frac{1}{(x^{2}+7x+4)^{2}}{\rm d}x=\frac{1}{1089}\left(2\sqrt{33}\log(7+\sqrt{33}+2x)-2\sqrt{33}\log(-7+\sqrt{33}-2x)-\frac{33(2x+7)}{x^{2}+7x+4}\right)+C$$

A. P.
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