Find the most general antiderivative of .
Note: Any arbitrary constants used must be an upper-case " C ".

Ask by Delgado Wagner. in the United States

Mar 30,2025

Question

Find the most general antiderivative of $$ f(u)=\frac{2 u^{4}-7 \sqrt{u}}{u^{2}} $$. Note: Any arbitrary constants used must be an upper-case " C ". $$ F(u)=\square $$

UpStudy AI · Accepted Answer

We start with the function
$$
f(u)=\frac{2u^4-7\sqrt{u}}{u^2}.
$$

**Step 1. Simplify the expression**

Divide each term in the numerator by $$ u^2 $$:
$$
f(u)=\frac{2u^4}{u^2}-\frac{7\sqrt{u}}{u^2}=2u^{4-2}-7u^{\frac{1}{2}-2}=2u^2-7u^{-\frac{3}{2}}.
$$

**Step 2. Find the antiderivative term-by-term**

We need to compute
$$
\int \left(2u^2-7u^{-\frac{3}{2}}\right)du.
$$

1. For the term $$ 2u^2 $$:
   $$
   \int 2u^2\, du = 2 \int u^2\, du = 2\left(\frac{u^3}{3}\right)=\frac{2u^3}{3}.
   $$

2. For the term $$ -7u^{-\frac{3}{2}} $$:
   
   We use the power rule for integration, which is valid for any exponent $$ n\neq -1 $$:
   $$
   \int u^n\,du=\frac{u^{n+1}}{n+1}+C.
   $$
   In our case, $$ n=-\frac{3}{2} $$, thus:
   $$
   \int u^{-\frac{3}{2}}\,du=\frac{u^{-\frac{3}{2}+1}}{-\frac{3}{2}+1} = \frac{u^{-\frac{1}{2}}}{-\frac{1}{2}}=-2u^{-\frac{1}{2}}.
   $$
   Multiplying by $$-7$$:
   $$
   -7\int u^{-\frac{3}{2}}\,du = -7(-2u^{-\frac{1}{2}})=14u^{-\frac{1}{2}}.
   $$

**Step 3. Combine the results and add an arbitrary constant**

The most general antiderivative is:
$$
F(u)=\frac{2u^3}{3}+14u^{-\frac{1}{2}}+C.
$$

Thus, the final answer is:
$$
F(u)=\frac{2u^3}{3}+14u^{-\frac{1}{2}}+C.
$$
$$ F(u) = \frac{2u^3}{3} + 14u^{-\frac{1}{2}} + C $$

Find the most general antiderivative of .
Note: Any arbitrary constants used must be an upper-case " C ".

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Find the most general antiderivative of . Note: Any arbitrary constants used must be an upper-case " C ".