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I have tried to solve this integral $$\int\frac{x^2}{(x+6)^6}\,\mathrm dx$$ by using partial fractions.

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    Hint: try $x+6=t$ – 2017-01-20

2 Answers 2

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Try substitution. Let $u = x + 6$, so $x = u-6$ and $du = dx$. Then we get

$$\int \frac{(u-6)^2}{u^6}du = \int \frac{u^2-12u+36}{u^6} du = \int \frac{1}{u^4} - \frac{12}{u^5} + \frac{36}{u^6} du$$

You should be able to complete the integral from there - and don't forget to plug back in for $x$!

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    It's still partial fractions – 2017-01-20
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    @Renascence_5. It is ultimately in multiple fractions, but I assume the OP meant that they didn't want to use the technique of partial fraction decomposition. I used substitution, not partial fraction decomposition. – 2017-01-20
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HINT:

Write $x^2$ as

$$x^2=(x+6)^2-12(x+6)+36$$

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    Taylor Series. ${}{}{}{}{}{}{}{}$ – 2017-01-20