Blog: `int x(2x + 5)^8 dx` Evaluate the indefinite integral.

Monday, June 13, 2011

$\displaystyle\int{x}{{\left({2}{x}+{5}\right)}}^{{8}}{\left.{d}{x}\right.}$ Evaluate the indefinite integral.

Since it would be very hard to raise to the 8th power the binomial 2x + 5, you need to use the following substitution $\displaystyle{2}{x}+{5}={u}$ , such that:

$\displaystyle{2}{x}+{5}={u}\Rightarrow{2}{\left.{d}{x}\right.}={d}{u}\Rightarrow{\left.{d}{x}\right.}=\frac{{{d}{u}}}{{2}}$

$\displaystyle{x}=\frac{{{u}-{5}}}{{2}}$

$\displaystyle\int{x}\cdot{{\left({2}{x}+{5}\right)}}^{{8}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{4}}\right)}\int{\left({u}-{5}\right)}\cdot{{u}}^{{8}}{d}{u}$

$\displaystyle{\left(\frac{{1}}{{4}}\right)}\int{\left({u}-{5}\right)}\cdot{{u}}^{{8}}{d}{u}={\left(\frac{{1}}{{4}}\right)}\int{{u}}^{{9}}{d}{u}-{\left(\frac{{1}}{{4}}\right)}\int{5}{{u}}^{{8}}{d}{u}$

$\displaystyle{\left(\frac{{1}}{{4}}\right)}\int{\left({u}-{5}\right)}\cdot{{u}}^{{8}}{d}{u}=\frac{{{{u}}^{{10}}}}{{40}}-\frac{{{5}{{u}}^{{9}}}}{{36}}+{c}$

Replacing back $\displaystyle{2}{x}+{5}$ for u yields:

$\displaystyle\int{x}\cdot{{\left({2}{x}+{5}\right)}}^{{8}}{\left.{d}{x}\right.}=\frac{{{{\left({2}{x}+{5}\right)}}^{{10}}}}{{40}}-\frac{{{5}{{\left({2}{x}+{5}\right)}}^{{9}}}}{{36}}+{c}$

Hence, evaluating the indefinite integral, yields $\displaystyle\int{x}\cdot{{\left({2}{x}+{5}\right)}}^{{8}}{\left.{d}{x}\right.}=\frac{{{{\left({2}{x}+{5}\right)}}^{{10}}}}{{40}}-\frac{{{5}{{\left({2}{x}+{5}\right)}}^{{9}}}}{{36}}+{c}.$

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Monday, June 13, 2011

$\displaystyle\int{x}{{\left({2}{x}+{5}\right)}}^{{8}}{\left.{d}{x}\right.}$ Evaluate the indefinite integral.

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