Question

find \\( \int \left(9x^3 + 7x^2 - 9x + 5\
ight) dx \\).
\\( \int \left(9x^3 + 7x^2 - 9x + 5\
ight) dx = \square \\)

Explanation:

Step1: Apply Sum Rule of Integration

The integral of a sum is the sum of integrals: $\int (9x^3 + 7x^2 - 9x + 5) dx=\int 9x^3 dx+\int 7x^2 dx+\int (-9x) dx+\int 5 dx$

Step2: Apply Constant Multiple Rule

Factor out constants from each integral: $9\int x^3 dx + 7\int x^2 dx- 9\int x dx+5\int 1 dx$

Step3: Apply Power Rule for Integration

The power rule is $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ (for $n
eq - 1$) and $\int 1 dx=x + C$.

• For $\int x^3 dx$: $n = 3$, so $\int x^3 dx=\frac{x^{3 + 1}}{3+1}=\frac{x^4}{4}$
• For $\int x^2 dx$: $n=2$, so $\int x^2 dx=\frac{x^{2 + 1}}{2+1}=\frac{x^3}{3}$
• For $\int x dx$: $n = 1$, so $\int x dx=\frac{x^{1+1}}{1 + 1}=\frac{x^2}{2}$
• For $\int 1 dx$: it is $x$

Step4: Combine Terms

Multiply back the constants and combine:
$9\times\frac{x^4}{4}+7\times\frac{x^3}{3}-9\times\frac{x^2}{2}+5\times x+C=\frac{9}{4}x^4+\frac{7}{3}x^3-\frac{9}{2}x^2 + 5x+C$

Answer:

$\frac{9}{4}x^4+\frac{7}{3}x^3-\frac{9}{2}x^2 + 5x+C$