Blog: `int x^2 sqrt(x^3 + 1) dx, u = x^3 + 1` Evaluate the integral by making the given substitution.

Sunday, April 12, 2015

$\displaystyle\int{{x}}^{{2}}\sqrt{{{{x}}^{{3}}+{1}}}{\left.{d}{x}\right.},{u}={{x}}^{{3}}+{1}$ Evaluate the integral by making the given substitution.

You need to evaluate the indefinite integral by performing the indicated substitution $\displaystyle{u}={{x}}^{{3}}+{1}$ , such that:

$\displaystyle{u}={{x}}^{{3}}+{1}\Rightarrow{d}{u}={3}{{x}}^{{2}}{\left.{d}{x}\right.}\Rightarrow{{x}}^{{2}}{\left.{d}{x}\right.}=\frac{{{d}{u}}}{{3}}$

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

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

$\displaystyle{\left(\frac{{1}}{{3}}\right)}\cdot\int\sqrt{{u}}{d}{u}={\left(\frac{{1}}{{3}}\right)}\frac{{{{u}}^{{\frac{{1}}{{2}}+{1}}}}}{{\frac{{1}}{{2}}+{1}}}+{c}$

$\displaystyle{\left(\frac{{1}}{{3}}\right)}\cdot\int\sqrt{{u}}{d}{u}={\left(\frac{{2}}{{9}}\right)}{\left({{u}}^{{\frac{{3}}{{2}}}}\right)}+{c}$

Replacing back $\displaystyle{{x}}^{{3}}+{1}$ for u yields:

$\displaystyle\int{{x}}^{{2}}\sqrt{{{{x}}^{{3}}+{1}}}{\left.{d}{x}\right.}={\left(\frac{{2}}{{9}}\right)}{\left({{\left({{x}}^{{3}}+{1}\right)}}^{{\frac{{3}}{{2}}}}\right)}+{c}$

Hence, evaluating the indefinite integral yields $\displaystyle\int{{x}}^{{2}}\sqrt{{{{x}}^{{3}}+{1}}}{\left.{d}{x}\right.}={\left(\frac{{2}}{{9}}\right)}{\left({{x}}^{{3}}+{1}\right)}\sqrt{{{{x}}^{{3}}+{1}}}+{c}$

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Sunday, April 12, 2015

$\displaystyle\int{{x}}^{{2}}\sqrt{{{{x}}^{{3}}+{1}}}{\left.{d}{x}\right.},{u}={{x}}^{{3}}+{1}$ Evaluate the integral by making the given substitution.

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