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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<math>\begin{align} | <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 | \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 | ||
\end{align}</math> | |||
Revision as of 19:20, 2 September 2022
![{\displaystyle {\begin{aligned}\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\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/cae5cc284ce45518a7c5eda74b1f2146b537c547)