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
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[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
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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]
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