Blog: `int (dx)/(5 - 3x)` Evaluate the indefinite integral.

Monday, November 28, 2016

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{5}-{3}{x}}}$ Evaluate the indefinite integral.

You need to use the following substitution 5 - 3x = t, such that:

$\displaystyle{5}-{3}{x}={t}\Rightarrow-{3}{\left.{d}{x}\right.}={\left.{d}{t}\right.}\Rightarrow{\left.{d}{x}\right.}=-\frac{{{\left.{d}{t}\right.}}}{{3}}$

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{5}-{3}{x}}}=-{\left(\frac{{1}}{{3}}\right)}\int\frac{{{\left.{d}{t}\right.}}}{{t}}$

$\displaystyle-{\left(\frac{{1}}{{3}}\right)}\int\frac{{{\left.{d}{t}\right.}}}{{t}}=-{\left(\frac{{1}}{{3}}\right)}{\ln}{\left|{t}\right|}+{c}$

Replacing back 5 - 3x for t yields:

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{5}-{3}{x}}}=-{\left(\frac{{1}}{{3}}\right)}{\ln}{\left|{5}-{3}{x}\right|}+{c}$

Hence, evaluating the indefinite integral, yields $\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{5}-{3}{x}}}=-{\left(\frac{{1}}{{3}}\right)}{\ln}{\left|{5}-{3}{x}\right|}+{c}.$

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Monday, November 28, 2016

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{5}-{3}{x}}}$ Evaluate the indefinite integral.

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