Answer: [tex]\frac{3}{4}(2+t)^{4/3}+C\\\\[/tex]
where C is a constant.
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Explanation:
Apply u-substitution
[tex]u = 2+t\\\\\frac{du}{dt} = 1\\\\du = dt[/tex]
So,
[tex]\displaystyle \int \sqrt[3]{2+t}\ \ dt = \int (2+t)^{1/3}\ dt\\\\\displaystyle \int \sqrt[3]{2+t}\ \ dt = \int (u)^{1/3} \ du\\\\\displaystyle \int \sqrt[3]{2+t}\ \ dt = \frac{1}{1+1/3}(u)^{1+1/3}+C\\\\[/tex]
[tex]\displaystyle \int \sqrt[3]{2+t}\ \ dt = \frac{1}{4/3}(u)^{4/3}+C\\\\\displaystyle \int \sqrt[3]{2+t}\ \ dt = \frac{3}{4}(u)^{4/3}+C\\\\\displaystyle \int \sqrt[3]{2+t}\ \ dt = \frac{3}{4}(2+t)^{4/3}+C\\\\[/tex]
Don't forget about the plus C at the end.
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The rule used in step 3 is
[tex]\displaystyle \int x^n \ dx = \frac{1}{n+1}x^{n+1}+C[/tex]
To verify the answer, apply the derivative to both sides and you should find that
[tex]\displaystyle \frac{d}{dt}\int \sqrt[3]{2+t}\ \ dt = \frac{d}{dt}\left[\frac{3}{4}(2+t)^{4/3}+C\right] = \sqrt[3]{2+t}\\\\[/tex]
I'm skipping a bit of steps.
This is an example of the fundamental theorem of calculus to tie together the inverse operations of derivatives vs integrals (aka antiderivatives).
WolframAlpha is a tool you can use to verify the answer.
