Calc1.72

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$\int \frac{x}{1-x^2}\,dx\,$

$u=1-x^2\,$

$du=-2x\,dx\,$

$\int \frac{x}{1-x^2}\,dx=-\frac{1}{2}\int \frac{du}{u}=-\frac{1}{2}\ln |1-x^2|+C\,$

Click here to see this integral done by trigonometric substition.
Click here to see this integral done by partial fractions.

Main Page : Calculus

Calc1.72

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