5.5 The Substitution Rule/43 - Revision history

Kimberlyr70044@students.laalliance.org at 19:20, 2 September 2022

2022-09-02T19:20:56Z

← Older revision		Revision as of 19:20, 2 September 2022
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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\end{align}</math>
	\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>

Kimberlyr70044@students.laalliance.org at 19:20, 2 September 2022

2022-09-02T19:20:11Z

← Older revision		Revision as of 19:20, 2 September 2022
Line 1:		Line 1:
	<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
	~~<\math>~~\end{align}		\end{align}</math>

Kimberlyr70044@students.laalliance.org: Created page with " $\begin{align} \int\frac{1+x}{1+x^2}dx &=\int(\frac{1}{1+x^2}+\frac{x}{1+x^2})dx\\[2ex]&=\int\frac{1}{1+x^2}dx+\int\frac{x}{1+x^2}dx\\[2ex]&=tan^{-1}(x) + \frac{1}{2}\int\frac{1}{u}du\\[2ex]&= tan^{-1}(x)+\ln|{1+x}|+c <\math>\end{align}"$

2022-09-02T19:19:55Z

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}"

New page

<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}

5.5 The Substitution Rule/43 - Revision history

Kimberlyr70044@students.laalliance.org at 19:20, 2 September 2022

Kimberlyr70044@students.laalliance.org at 19:20, 2 September 2022

Kimberlyr70044@students.laalliance.org: Created page with "\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}"

Kimberlyr70044@students.laalliance.org: Created page with " $\begin{align} \int\frac{1+x}{1+x^2}dx &=\int(\frac{1}{1+x^2}+\frac{x}{1+x^2})dx\\[2ex]&=\int\frac{1}{1+x^2}dx+\int\frac{x}{1+x^2}dx\\[2ex]&=tan^{-1}(x) + \frac{1}{2}\int\frac{1}{u}du\\[2ex]&= tan^{-1}(x)+\ln|{1+x}|+c <\math>\end{align}"$