$\frac{1}{2}\int \left(y-3\right) \sqrt{y} \mathrm{dy}$

$\=\frac{1}{2}\int y^{3\/2} \mathit{\ⅆ}y-\frac{3}{2}\int \sqrt{y}\mathit{\ⅆ}y$

$\=\frac{1}{2} \frac{y^{5\/2}}{5\/2}-\frac{3}{2} \frac{y^{3\/2}}{3\/2}$

$\={\left(x^2\+3\right)}^{5\/2}\/5-{\left(x^2\+3\right)}^{3\/2}$

$\int \left(u^2-3\right) u^2 \mathrm{du}$

$\= \int u^4 \mathrm{du}-3\int u^2 \mathrm{du}$

$\=u^5\/5-u^3$

$\={\left(x^2\+3\right)}^{5\/2}\/5-{\left(x^2\+3\right)}^{3\/2}$

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$\int x^3\sqrt{x^2+3}\mathit{\ⅆ}x$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Integration Steps}\\ & & \int 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\mathrm{dx}=2u\mathrm{du}\\ & \text{◦}& \text{Isolate equation for}x\\ & & x=u\\ & \text{◦}& \text{Substitute the values back into the original}\\ & & \int x^3\sqrt{x^2+3}\ⅆx=\int \left(u^4-3u^2\right)\ⅆu\\ & & \text{This gives:}\\ & & \int \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 u^4\ⅆu+\int -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^n\ⅆu=\frac{u^{n+1}}{n+1}\\ & \text{◦}& \text{This means:}\\ & & \int u^4\ⅆu=\frac{u^{4+1}}{4+1}\\ & \text{◦}& \text{So,}\\ & & \int u^4\ⅆu=\frac{u^5}{5}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{u^5}{5}+\int -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 Cf\left(u\right)\ⅆu=C\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{u^5}{5}-3\left(\int 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^n\ⅆu=\frac{u^{n+1}}{n+1}\\ & \text{◦}& \text{This means:}\\ & & \int u^2\ⅆu=\frac{u^{2+1}}{2+1}\\ & \text{◦}& \text{So,}\\ & & \int u^2\ⅆu=\frac{u^3}{3}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{1}{5}{u}^5-u^3\\ \text{▫}& & \text{6. Revert change of variable}\\ & \text{◦}& \text{Variable we defined in step}1\\ & & x^2+3=u^2\\ & & \text{This gives:}\\ & & \frac{{\left(x^2+3\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{5}-{\left(x^2+3\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

Table 4.4.4(a)   Stepwise evaluation via the Context Panel's "Student Calculus1≻All Solution Steps" option

Table 4.4.4(b)   The substitution $y\=x^2\+3$ implemented in the Integration Methods tutor