Integrate 5³+17x-7 all over (x²+4)(x-1)dx

$$5^3+17x-7 = 125+17x-7 = 17x+118$$

So the integral becomes:

$$\int \frac{17x+118}{(x^2+4)(x-1)} \, dx$$

Since $(x^2+4)$ is an irreducible quadratic factor and $(x-1)$ is a linear factor, the decomposition takes the form:

$$\frac{17x+118}{(x^2+4)(x-1)} = \frac{Ax+B}{x^2+4} + \frac{C}{x-1}$$

Multiply both sides by $(x^2+4)(x-1)$ to clear the denominators:

$$17x+118 = (Ax+B)(x-1) + C(x^2+4)$$

Expand the right side:

$$17x+118 = Ax^2 - Ax + Bx - B + Cx^2 + 4C$$

Group terms by powers of $x$:

$$17x+118 = (A+C)x^2 + (-A+B)x + (-B+4C)$$

Equate the coefficients of corresponding powers of $x$ on both sides:
For $x^2$: $A+C = 0 \quad (*1)$
For $x$: $-A+B = 17 \quad (*2)$
For constant: $-B+4C = 118 \quad (*3)$

From $(*1)$, $A = -C$. Substitute this into $(*2)$:
$-(-C)+B = 17 \implies C+B = 17 \implies B = 17-C$.
Substitute $B = 17-C$ into $(*3)$:
$-(17-C)+4C = 118$
$-17+C+4C = 118$
$5C = 118+17$
$5C = 135$
$C = \frac{135}{5} = 27$

Now find A and B:
$A = -C = -27$.
$B = 17-C = 17-27 = -10$.

So the partial fraction decomposition is:

$$\frac{-27x-10}{x^2+4} + \frac{27}{x-1}$$

$$\int \left( \frac{-27x-10}{x^2+4} + \frac{27}{x-1} \right) dx = \int \frac{-27x}{x^2+4} \, dx - \int \frac{10}{x^2+4} \, dx + \int \frac{27}{x-1} \, dx$$

For the first integral, $\int \frac{-27x}{x^2+4} \, dx$:
Let $u = x^2+4$, then $du = 2x \, dx$, so $x \, dx = \frac{1}{2} du$.

$$\int \frac{-27}{u} \left(\frac{1}{2} du\right) = -\frac{27}{2} \int \frac{1}{u} du = -\frac{27}{2} \ln|u| = -\frac{27}{2} \ln(x^2+4)$$

(Since $x^2+4$ is always positive, we can remove the absolute value.)

For the second integral, $\int \frac{10}{x^2+4} \, dx$:
This is of the form $\int \frac{1}{x^2+a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right)$. Here $a^2=4$, so $a=2$.

$$10 \int \frac{1}{x^2+2^2} \, dx = 10 \cdot \frac{1}{2} \arctan\left(\frac{x}{2}\right) = 5 \arctan\left(\frac{x}{2}\right)$$

For the third integral, $\int \frac{27}{x-1} \, dx$:

$$27 \int \frac{1}{x-1} \, dx = 27 \ln|x-1|$$

$$-\frac{27}{2} \ln(x^2+4) - 5 \arctan\left(\frac{x}{2}\right) + 27 \ln|x-1| + C$$

The final answer is $\boxed{-\frac{27}{2} \ln(x^2+4) - 5 \arctan\left(\frac{x}{2}\right) + 27 \ln|x-1| + C}$.