From Example 4.4.4, if $u^2\=x^2\+3$, so that $x\=2\Rightarrow u\=\sqrt{7}$, and $x\=3\Rightarrow u\=\sqrt{12}$, then

From Example 4.4.4, making either the substitution $u^2\=x^2\+3$, or $y\=x^2\+3$, the indefinite integral of $x^3\sqrt{x^2\+3}$ is ${\left(x^2\+3\right)}^{5\/2}\/5-{\left(x^2\+3\right)}^{3\/2}$. Hence, the value of the definite integral is

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${\int }_2^3{x}^3\sqrt{x^2+3} \mathit{\ⅆ}x$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Integration Steps}\\ & & {\int }_2^3{x}^3\sqrt{x^2+3}\ⅆx\\ \text{▫}& & \text{1. Apply a change of variables to rewrite the integral in terms of}u\\ & \text{◦}& \text{Let}\\ & & x^2+3=u^2\\ & \text{◦}& \text{Differentiate both sides}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2+3\right)=\frac{\ⅆ}{\ⅆu}\left(u^2\right)\\ & \text{◦}& \text{Evaluate}\\ & & 2x\left[\right]=2u\left[\right]\\ & \text{◦}& \text{Isolate equation for}x\\ & & x=u\\ & \text{◦}& \text{Substitute the values back into the original}\\ & & {\int }_2^3{x}^3\sqrt{x^2+3}\ⅆx={\int }_{\sqrt{7}}^{\sqrt{12}}\left(u^4-3u^2\right)\ⅆu\\ & & \text{This gives:}\\ & & {\int }_{\sqrt{7}}^{\sqrt{12}}\left(u^4-3u^2\right)\ⅆu\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{sum}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{sum}}\text{rule}\\ & & \int \left(f\left(u\right)+g\left(u\right)\right)\ⅆu=\int f\left(u\right)\ⅆu+\int g\left(u\right)\ⅆu\\ & & f\left(u\right)=u^4\\ & & g\left(u\right)=-3u^2\\ & & \text{This gives:}\\ & & {\int }_{\sqrt{7}}^{\sqrt{12}}{u}^4\ⅆu+{\int }_{\sqrt{7}}^{\sqrt{12}}-3u^2\ⅆu\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{power}}\text{rule to the term}\int u^4\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule, for n}\text{≠}\text{-1}\\ & & \int u^{\left[\right]}\ⅆu=\left[\right]\\ & \text{◦}& \text{This means:}\\ & & \int u^4\ⅆu=\left[\right]\\ & \text{◦}& \text{So,}\\ & & \int u^4\ⅆu=\frac{u^5}{5}\\ & \text{◦}& \text{Apply limits of definite integral}\\ & & \left[\right]-\left(\left[\right]\right)\\ & & \text{We can rewrite the integral as:}\\ & & \frac{144\sqrt{12}}{5}-\frac{49\sqrt{7}}{5}+{\int }_{\sqrt{7}}^{\sqrt{12}}-3u^2\ⅆu\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int -3u^2\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int \left[\right]f\left(u\right)\ⅆu=\left[\right]\left(\int f\left(u\right)\ⅆu\right)\\ & \text{◦}& \text{This means:}\\ & & \int -3u^2\ⅆu=-3\left(\int u^2\ⅆu\right)\\ & & \text{We can rewrite the integral as:}\\ & & \frac{144\sqrt{12}}{5}-\frac{49\sqrt{7}}{5}-3\left({\int }_{\sqrt{7}}^{\sqrt{12}}{u}^2\ⅆu\right)\\ \text{▫}& & \text{5. Apply the}\mathbf{\text{power}}\text{rule to the term}\int u^2\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule, for n}\text{≠}\text{-1}\\ & & \int u^{\left[\right]}\ⅆu=\left[\right]\\ & \text{◦}& \text{This means:}\\ & & \int u^2\ⅆu=\left[\right]\\ & \text{◦}& \text{So,}\\ & & \int u^2\ⅆu=\frac{u^3}{3}\\ & \text{◦}& \text{Apply limits of definite integral}\\ & & \left[\right]-\left(\left[\right]\right)\\ & & \text{We can rewrite the integral as:}\\ & & \frac{84\sqrt{12}}{5}-\frac{14\sqrt{7}}{5}\end{array}$

Table 4.4.5(a)   Maple's annotated stepwise evaluation of the definite integral