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Using the Table of Integrals/Antiderivatives, calculate the integral of $y=\\simplify[unitFactor, unitPower,fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}}$.

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From the Table of Integrals we see that a function of the form \\[ f(x)=x^n \\] has the integral \\[ \\int x^n dx  =  \\frac{x^{n+1}}{n+1}+ c,\\]

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and \\[\\int kf(x) dx = k \\int f(x) dx.\\]

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Additionally, the integral of the sum or difference of two or more functions is equal to the sum or difference of the integrals of each function: \\[ \\int(f(x)\\pm g(x))\\, dx = \\int f(x)\\, dx  \\pm \\int g(x) \\, dx.\\]

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So, for the function

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\\[y=\\simplify[all,fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}},\\]

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the integral  is

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\\[ \\begin{split}\\simplify[all,fractionNumbers]{int({a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3},x)} = \\simplify[all,fractionNumbers]{{a_1}int(x^{b_1},x)+{a_2}int(x^{b_2},x)+{a_3}int(x^{b_3},x)} &\\,= \\simplify[all,fractionNumbers,!collectNumbers,!simplifyFractions,!noLeadingMinus]{{a_1}(x^({b_1+1})/{b_1+1}) +{a_2}(x^({b_2+1})/{b_2+1}) +{a_3}(x^({b_3+1})/{b_3+1})} + c,\\\\ \\\\&\\,= \\simplify[all,fractionNumbers]{{a_1/(b_1+1)}x^{b_1+1}+{a_2/(b_2+1)}x^{b_2+1}+{a_3/(b_3+1)}x^{b_3+1}}+c.\\end{split} \\]

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Note: You only need to put one $c$ term here, you do not need to put a separate constant term for each calculation. 

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