Hi,

I'm trying to integrate:

$\displaystyle 1/(ax - x^3)$

I've tried using partial fractions; and I've got (1/a) lnx and a^(-3/2) tanh(x/a).

Thanks for any help!

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- Feb 13th 2011, 11:52 AMAntIndefinite integral
Hi,

I'm trying to integrate:

$\displaystyle 1/(ax - x^3)$

I've tried using partial fractions; and I've got (1/a) lnx and a^(-3/2) tanh(x/a).

Thanks for any help! - Feb 13th 2011, 12:04 PMKrizalid
substitute $\displaystyle x=\dfrac 1t$ and the integral becomes $\displaystyle \displaystyle-\int{\frac{t}{a{{t}^{2}}-1}\,dt}=-\frac{1}{2a}\ln \left| 1-a{{t}^{2}} \right|+k,$ so back substitute and get $\displaystyle -\dfrac{1}{2a}\ln \left| \dfrac{{{x}^{2}}-a}{{{x}^{2}}} \right|+k.$