Calc1.70

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\int {\frac  {1}{(x^{2}+1)^{{{\frac  {3}{2}}}}}}\,dx\,

x=\tan \theta \,

dx=\sec ^{2}\theta \,d\theta \,

\int {\frac  {1}{(x^{2}+1)^{{{\frac  {3}{2}}}}}}\,dx=\int {\frac  {\sec ^{2}\theta }{(tan^{2}\theta +1)^{{{\frac  {3}{2}}}}}}d\theta =\int {\frac  {\sec ^{2}\theta }{\sec ^{3}\theta }}d\theta =\int \cos \theta \,d\theta =\sin \theta +C\,

We know x=\tan \theta \, so \sin \theta ={\frac  {x}{{\sqrt  {x^{2}+1}}}}\,. Therefore

\int {\frac  {1}{(x^{2}+1)^{{{\frac  {3}{2}}}}}}\,dx={\frac  {x}{{\sqrt  {x^{2}+1}}}}+C\,


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