Division yields
[tex]\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}[/tex]
Now for partial fractions: you're looking for constants a, b, and c such that
[tex]\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}[/tex]
[tex]\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a[/tex]
which gives a + b = 2, c = 0, and 2a = -7, so that a = -7/2 and b = 11/2. Then
[tex]\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}[/tex]
Now, in the integral we get
[tex]\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx[/tex]
The first two terms are trivial to integrate. For the third, substitute y = x ² + 2 and dy = 2x dx to get
[tex]\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}[/tex]
