Is This Integral More Challenging Than It Appears?

[Nea]

Can you help me to solve this integral?
$$\int {\frac{{x^3 + 1}}{{x(x^3 - 8)}}} dx$$

 

[Muzza]

Partial fractions? x^3 - 8 = x^3 - 2^3 is easy to factor.

 

[arildno]

Remember that $8=2^{3}$
Hence, we have:
$x^{3}-8=(x-2)(x^{2}+2x+4)$
Use partial fractions decomposition; it might help to note that:
$$x^{2}+2x+4=(x+1)^{2}+3=3((\frac{x+1}{\sqrt{3}})^{2}+1)$$

 

[Orion1]

Formula for integration by parts by Substitution Rule:
$$\int u dv = uv - \int v du$$

$$u = x^3 + 1 \; \; \; dv = \frac{dx}{x^4-8x}$$

$$du = 3x^2 dx \; \; \; v = \left[\frac{\ln \left(x^3 - 8\right)}{24} - \frac{\ln\left(x\right)}{8}\right]$$

$$\int \left(x^3 + 1\right)\left(\frac{1}{x^4-8x}\right)dx = \left(x^3 + 1\right)\left[\frac{\ln \left(x^3 - 8\right)}{24} - \frac{\ln\left(x\right)}{8}\right] - \int \left[\frac{\ln \left(x^3 - 8\right)}{24} - \frac{\ln\left(x\right)}{8}\right]\left(3x^2\right)dx$$

 

[VietDao29]

...$$\int \left(x^3 + 1\right)\left(\frac{1}{x^4-8x}\right)dx = \left(x^3 + 1\right)\left[\frac{\ln \left(x^3 - 8\right)}{24} - \frac{\ln\left(x\right)}{8}\right] - \int \left[\frac{\ln \left(x^3 - 8\right)}{24} - \frac{\ln\left(x\right)}{8}\right]\left(3x^2\right)dx$$

Orion1, do you think that the latter integral look much more complicated than the former one??

 

[Orion1]

much more complicated than the former one??

Affirmative
$$\int \left(x^3 + 1\right)\left(\frac{1}{x^4-8x}\right)dx = \frac{1}{8} \left[ \left(x^3 + 1\right)\left[\frac{\ln \left(x^3 - 8\right)}{3} - \ln\left(x\right)\right] - \int \left[ \ln \left(x^3 - 8\right) - 3 \ln\left(x\right)\right]\left(x^2\right)dx\right]$$