Question

find the most general antiderivative of the function. (check your answer by differentiation. use c for the constant of the antiderivative.) g(u)=2u + \sqrt6{u}

Explanation:

Step1: Recall antiderivative rules

The antiderivative of a sum of functions is the sum of their antiderivatives. Also, the antiderivative of $au^n$ (where $a$ is a constant and $n
eq - 1$) is $\frac{a}{n + 1}u^{n+1}+C$.

Step2: Find antiderivative of $2u$

For the function $2u$, using the power - rule for antiderivatives with $a = 2$ and $n = 1$, we have $\int2u\ du=2\times\frac{1}{1 + 1}u^{1+1}=u^{2}$.

Step3: Find antiderivative of $\sqrt[6]{u}$

Rewrite $\sqrt[6]{u}=u^{\frac{1}{6}}$. Using the power - rule for antiderivatives with $a = 1$ and $n=\frac{1}{6}$, we get $\int u^{\frac{1}{6}}\ du=\frac{1}{\frac{1}{6}+1}u^{\frac{1}{6}+1}=\frac{1}{\frac{7}{6}}u^{\frac{7}{6}}=\frac{6}{7}u^{\frac{7}{6}}$.

Step4: Combine antiderivatives

The antiderivative of $g(u)=2u+\sqrt[6]{u}$ is $\int(2u + u^{\frac{1}{6}})\ du=u^{2}+\frac{6}{7}u^{\frac{7}{6}}+C$.

Answer:

$u^{2}+\frac{6}{7}u^{\frac{7}{6}}+C$