** Final answer to the problem

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** Step-by-step Solution **

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- Integrate by partial fractions
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- Weierstrass Substitution
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We can factor the polynomial $27x^3+135x^2+225x+125$ using the rational root theorem, which guarantees that for a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ there is a rational root of the form $\pm\frac{p}{q}$, where $p$ belongs to the divisors of the constant term $a_0$, and $q$ belongs to the divisors of the leading coefficient $a_n$. List all divisors $p$ of the constant term $a_0$, which equals $125$

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$1, 5, 25, 125$

Learn how to solve factorization problems step by step online. Factor the expression 27x^3+135x^2225x+125. We can factor the polynomial 27x^3+135x^2+225x+125 using the rational root theorem, which guarantees that for a polynomial of the form a_nx^n+a_{n-1}x^{n-1}+\dots+a_0 there is a rational root of the form \pm\frac{p}{q}, where p belongs to the divisors of the constant term a_0, and q belongs to the divisors of the leading coefficient a_n. List all divisors p of the constant term a_0, which equals 125. Next, list all divisors of the leading coefficient a_n, which equals 27. The possible roots \pm\frac{p}{q} of the polynomial 27x^3+135x^2+225x+125 will then be. Trying all possible roots, we found that -\frac{5}{3} is a root of the polynomial. When we evaluate it in the polynomial, it gives us 0 as a result.

** Final answer to the problem

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