Question

determine the following indefinite integral.

$$ int \frac { 3 x ^ { 5 } + 6 x ^ { 4 } } { x ^ { 3 } } d x $$

$$ int \frac { 3 x ^ { 5 } + 6 x ^ { 4 } } { x ^ { 3 } } d x = $$

Explanation:

Step1: Simplify the integrand

First, we simplify the fraction \(\frac{3x^5 + 6x^4}{x^3}\) by dividing each term in the numerator by \(x^3\). Using the rule of exponents \( \frac{x^m}{x^n}=x^{m - n}\), we get:
\(\frac{3x^5}{x^3}+\frac{6x^4}{x^3}=3x^{5 - 3}+6x^{4 - 3}=3x^2 + 6x\)

Step2: Integrate term - by - term

Now we integrate the simplified function \(3x^2+6x\) with respect to \(x\). We use the power rule for integration \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)) for each term.

• For the term \(3x^2\):

\(\int3x^2dx = 3\times\frac{x^{2 + 1}}{2+1}=3\times\frac{x^3}{3}=x^3\)

• For the term \(6x\):

\(\int6xdx=6\times\frac{x^{1+1}}{1 + 1}=6\times\frac{x^2}{2}=3x^2\)

Step3: Combine the results and add the constant of integration

We add the results of the two integrations and include the constant of integration \(C\) (since it's an indefinite integral). So the integral is \(x^3+3x^2 + C\)

Answer:

\(x^{3}+3x^{2}+C\) (where \(C\) is the constant of integration)