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**Hint:**In mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. The process of finding integrals is called integration. The integration of the two functions multiplied with each other can be evaluated by using integration by parts.

**Complete step by step answer:**

Here we have the function as \[\int{(x+3)(x-1)}dx\]

Integration by parts is given as

\[\int{(u)(v)dx=(u)\int{(v)dx-\int{\left( \dfrac{d(u)}{dx}\int{(v)dx} \right)dx}}}\]

Where \[u\] and \[v\] are two variables/ functions

The formula can also be seen as, “first term multiplied by integration of second term, minus the integration of differentiation of first term multiplied with the integration of second term.”

Now here we can see that the given function is a multiplication of two functions that is \[(x+3)\] and \[(x-1)\], Hence we can say \[u=(x+3)\] and \[v=(x-1)\]

Applying integration by parts we get,

\[\int{(x+3)(x-1)dx=(x+3)\int{(x-1)dx-\int{\left( \dfrac{d(x+3)}{dx}\int{(x-1)dx} \right)dx}}}\]

As you know that \[\dfrac{d(x+3)}{dx}=1\] and \[\int{(x-1)=\dfrac{{{(x-1)}^{2}}}{2}}\]

(Using integration property \[\int{{{(x+a)}^{n}}=\dfrac{{{(x+a)}^{n+1}}}{n+1}}\])

Substituting values of integral and differentiation we get,

\[\int{(x+3)(x-1)dx=(x+3)\left( \dfrac{{{(x-1)}^{2}}}{2} \right)-\int{\left( (1)\left( \dfrac{{{(x-1)}^{2}}}{2} \right) \right)}}\]

\[\Rightarrow \int{(x+3)(x-1)dx=(x+3)\left( \dfrac{{{(x-1)}^{2}}}{2} \right)-\int{\left( \dfrac{{{(x-1)}^{2}}}{2} \right)}}\]

Taking half common from the integral,

\[\Rightarrow \int{(x+3)(x-1)dx=(x+3)\left( \dfrac{{{(x-1)}^{2}}}{2} \right)-\dfrac{1}{2}\int{{{\left( x-1 \right)}^{2}}}}\]

Solving the integral using property \[\int{{{(x+a)}^{n}}=\dfrac{{{(x+a)}^{n+1}}}{n+1}}\] we get

\[\Rightarrow \int{(x+3)(x-1)dx=(x+3)\left( \dfrac{{{(x-1)}^{2}}}{2} \right)-\dfrac{1}{2}\left( \dfrac{{{(x-1)}^{3}}}{3} \right)}\]

Taking the common term outside we get,

\[\Rightarrow \dfrac{{{(x-1)}^{2}}}{2}\left( (x+3)-\dfrac{(x-1)}{3} \right)\]

Solving the internal of the bracket by taking 3 as LCM we get,

\[\Rightarrow \dfrac{{{(x-1)}^{2}}}{2}\left( \dfrac{3(x+3)-(x-1)}{3} \right)\]

\[\Rightarrow \dfrac{{{(x-1)}^{2}}}{2}\left( \dfrac{3x+9-x+1}{3} \right)\]

\[\Rightarrow \dfrac{{{(x-1)}^{2}}}{2}\left( \dfrac{2x+10}{3} \right)\]

Now we will take 2 common from the numerator and cancel it from the denominator.

\[\Rightarrow {{(x-1)}^{2}}\left( \dfrac{x+5}{3} \right)\]

Hence, \[\int{(x+3)(x-1)}={{(x-1)}^{2}}\left( \dfrac{x+5}{3} \right)+c\], (where c constant)

Since we need to evaluate it over [1, 3] we will apply the limits on it to get the final answer.

\[\Rightarrow {{(3-1)}^{2}}\left( \dfrac{3+5}{3} \right)-{{(1-1)}^{2}}\left( \dfrac{1+5}{3} \right)\]

\[\Rightarrow {{(2)}^{2}}\left( \dfrac{8}{3} \right)-{{(0)}^{2}}\left( \dfrac{6}{3} \right)\]

\[\Rightarrow \dfrac{32}{3}\]

**Hence \[\int{(x+3)(x-1)}\] over [1, 3] is \[\dfrac{32}{3}\].**

**Note:**In integration by parts, we have learned when the product of two functions are given to us then we apply the required formula. The integral of the two functions is taken, by considering the left term as first function and second term as the second function. We should keep in mind that the second term selected should be easily integrable and the first function is chosen in such a way that the derivative of the function could be easily integrated. Usually, the preference order of this rule is based on some functions such as Inverse, Algebraic, Logarithm, Trigonometric, Exponent. This method is called the ILATE rule.

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