Blog: `int x^3 (2 + x^4)^5 dx, u = 2 + x^4` Evaluate the integral by making the given substitution.

Tuesday, November 25, 2014

$\displaystyle\int{{x}}^{{3}}{{\left({2}+{{x}}^{{4}}\right)}}^{{5}}{\left.{d}{x}\right.},{u}={2}+{{x}}^{{4}}$ Evaluate the integral by making the given substitution.

You need to evaluate the indefinite integral by performing the indicated substitution $\displaystyle{u}={2}+{{x}}^{{4}}$ , such that:

$\displaystyle{u}={2}+{{x}}^{{4}},\Rightarrow{d}{u}={4}{{x}}^{{3}}{\left.{d}{x}\right.}\Rightarrow{{x}}^{{3}}{\left.{d}{x}\right.}=\frac{{{d}{u}}}{{4}}$

$\displaystyle\int{{x}}^{{3}}\cdot{{\left({2}+{{x}}^{{4}}\right)}}^{{5}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{4}}\right)}\cdot\int{{u}}^{{5}}{d}{u}$

Using the formula $\displaystyle\int{{u}}^{{n}}{d}{u}=\frac{{{{u}}^{{{n}+{1}}}}}{{{n}+{1}}}+{c}$ yields

$\displaystyle{\left(\frac{{1}}{{4}}\right)}\cdot\int{{u}}^{{5}}{d}{u}={\left(\frac{{1}}{{4}}\right)}\frac{{{{u}}^{{{5}+{1}}}}}{{{5}+{1}}}+{c}$

$\displaystyle{\left(\frac{{1}}{{4}}\right)}\cdot\int{{u}}^{{5}}{d}{u}={\left(\frac{{1}}{{24}}\right)}{\left({{u}}^{{6}}\right)}+{c}$

Replacing back $\displaystyle{2}+{{x}}^{{4}}$ for u yields:

$\displaystyle\int{{x}}^{{3}}\cdot{{\left({2}+{{x}}^{{4}}\right)}}^{{5}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{24}}\right)}\cdot{{\left({2}+{{x}}^{{4}}\right)}}^{{6}}+{c}$

Hence, evaluating the indefinite integral yields $\displaystyle\int{{x}}^{{3}}\cdot{{\left({2}+{{x}}^{{4}}\right)}}^{{5}}{\left.{d}{x}\right.}=\frac{{{{\left({2}+{{x}}^{{4}}\right)}}^{{6}}}}{{24}}+{c}$

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Tuesday, November 25, 2014

$\displaystyle\int{{x}}^{{3}}{{\left({2}+{{x}}^{{4}}\right)}}^{{5}}{\left.{d}{x}\right.},{u}={2}+{{x}}^{{4}}$ Evaluate the integral by making the given substitution.

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