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Evaluate the indefinite integral:
[tex]\large\begin{array}{l} \mathtt{\displaystyle\int\! x^3\cdot \sqrt{x^2+1}\,dx} \end{array}[/tex]
[tex]\large\begin{array}{l} \texttt{Rewrite it conveniently:}\\\\ \mathtt{=\displaystyle\int\!x^2\cdot x\cdot \sqrt{x^2+1}\,dx}\\\\ \mathtt{=\displaystyle\int\!x^2\cdot \sqrt{x^2+1}\cdot x\,dx\qquad\quad(i)} \end{array}[/tex]
[tex]\large\begin{array}{l} \texttt{Instead of using a trigonometric substitution, just}\\\texttt{make a regular u-substitution for this one:}\\\\ \begin{array}{lclcl} \mathtt{u=\sqrt{x^2+1}}&~\Rightarrow~&\mathtt{u^2=x^2+1}&~\Rightarrow~&\mathtt{x^2=u^2-1}\\\\ &&&&\mathtt{\diagup\!\!\!\! 2x\,dx=\diagup\!\!\!\! 2u\,du}\\\\ &&&&\mathtt{x\,dx=u\,du} \end{array} \end{array}[/tex]
[tex]\large\begin{array}{l} \texttt{so the integral (i) becomes}\\\\ \mathtt{=\displaystyle\int\!(u^2-1)\cdot u\cdot u\,du}\\\\ \mathtt{=\displaystyle\int\!(u^2-1)\cdot u^2\,du}\\\\ \mathtt{=\displaystyle\int\!(u^4-u^2)\,du\qquad\quad(ii)} \end{array}[/tex]
[tex]\large\begin{array}{l} \texttt{Now, evaluate (ii) by applying the power rule:}\\\\ \mathtt{=\dfrac{u^{4+1}}{4+1}-\dfrac{u^{2+1}}{2+1}+C}\\\\ \mathtt{=\dfrac{u^5}{5}-\dfrac{u^3}{3}+C} \end{array}[/tex]
[tex]\large\begin{array}{l} \texttt{Last, substitute back for }\mathtt{u=\sqrt{x^2+1}}\texttt{ and then}\\\texttt{you get}\\\\ \mathtt{=\dfrac{\big(\!\sqrt{x^2+1}\big)^5}{5}-\dfrac{\big(\!\sqrt{x^2+1}\big)^3}{3}+C}\\\\ \mathtt{=\dfrac{\big[(x^2+1)^{1/2}\big]^5}{5}-\dfrac{\big[(x^2+1)^{1/2}\big]^3}{3}+C}\\\\ \mathtt{=\dfrac{(x^2+1)^{5/2}}{5}-\dfrac{(x^2+1)^{3/2}}{3}+C} \end{array}[/tex]
[tex]\large\begin{array}{l} \therefore~~\boxed{\begin{array}{c}\mathtt{\displaystyle\int\!x^3\cdot\sqrt{x^2+1}\,dx=\dfrac{1}{5}\,(x^2+1)^{5/2}-\dfrac{1}{3}\,(x^2+1)^{3/2}+C} \end{array}}\qquad\checkmark \end{array}[/tex]
[tex]\large\texttt{I hope this helps. =)}[/tex]
Tags: indefinite integral product polynomial square root sqrt trigonometric trig substitution differential integral calculus
