5.5 The Substitution Rule/43: Difference between revisions
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(Created page with "<math>\begin{align} \int\frac{1+x}{1+x^2}dx &=\int(\frac{1}{1+x^2}+\frac{x}{1+x^2})dx\\[2ex]&=\int\frac{1}{1+x^2}dx+\int\frac{x}{1+x^2}dx\\[2ex]&=tan^{-1}(x) + \frac{1}{2}\int\frac{1}{u}du\\[2ex]&= tan^{-1}(x)+\ln|{1+x}|+c <\math>\end{align}") |
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Revision as of 19:19, 2 September 2022
<math>\begin{align} \int\frac{1+x}{1+x^2}dx &=\int(\frac{1}{1+x^2}+\frac{x}{1+x^2})dx\\[2ex]&=\int\frac{1}{1+x^2}dx+\int\frac{x}{1+x^2}dx\\[2ex]&=tan^{-1}(x) + \frac{1}{2}\int\frac{1}{u}du\\[2ex]&= tan^{-1}(x)+\ln|{1+x}|+c <\math>\end{align}