Calc1.68

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

We could do this integral with partial fractions, but for instructive purposes let's use the method of trigonometric substitution.

x=\sin \theta \,

dx=\cos \theta \,d\theta \,

\int {\frac  {x}{1-x^{2}}}\,dx=\int {\frac  {\sin \theta \cos \theta }{1-\sin ^{2}\theta }}d\theta =\int {\frac  {\sin \theta \cos \theta }{\cos ^{2}\theta }}d\theta =\int {\frac  {\sin \theta }{\cos \theta }}d\theta \,

At this point, we have to do a substitution again.

u=\cos \theta \,

du=-\sin \theta \,d\theta \,

\int {\frac  {\sin \theta }{\cos \theta }}d\theta =-\int {\frac  {du}{u}}=-\ln |u|+C=-\ln |\cos \theta |+C\,

Now, we know x=\sin \theta and we need to know the value of \cos \theta . Using the well known trig identity, \sin ^{2}x+\cos ^{2}x=1, we get that \cos \theta ={\sqrt  {1-x^{2}}}. So

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

Click here to see this integral done by partial fractions.
Click here to see this integral done by substitution.


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