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[shizuo]

\(\displaystyle{ \int \ \frac{dx}{x^3+x}}\)

[luka52]

\(\displaystyle{ \int \frac{\mbox{d}x}{x^3 +x} = t ft( \frac{1}{x} - \frac{x}{1+x^2} \right) = \ln |x| - \frac{1}{2} \ln |1+x^2| + C}\)

[shizuo]

skąd mamy \(\displaystyle{ \frac{1}{2}}\) ?

[luka52]

\(\displaystyle{ \frac{x}{1+x^2} = \frac{1}{2} ft( \frac{2x}{1+x^2} \right)}\)
A \(\displaystyle{ (1+x^2)' = 2x}\) stąd \(\displaystyle{ \int \frac{x \, dx}{1+x^2} = \frac{1}{2} \ln |1+x^2| +C}\)

[shizuo]

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