Calc1.73

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

Since 1-x^{2}=(1-x)(1+x), we can use the method of partial fractions to compute this integral.

{\frac  {x}{1-x^{2}}}={\frac  {A}{1-x}}+{\frac  {B}{1+x}}\Rightarrow x=A(1+x)+B(1-x)\,

Since this is true for all values of x, pick values which will cause one of A or B to cancel out to find the other.

Let x=1\,, then A={\frac  {1}{2}}

Let x=-1\,, then B=-{\frac  {1}{2}}

Thus, the integral is split up into two integrals

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

Click here to see this integral done by substitution.
Click here to see this integral done by trigonometric substition.


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