find the integral.\n int (9x^{2}-3x + 4)mathrm{d}x\n int (9x^{2}-3x + 4)mathrm{d}x=square

Explanation:

Step1: Apply integral rules

Use $\int (f(x)+g(x)+h(x))dx=\int f(x)dx+\int g(x)dx+\int h(x)dx$ and $\int ax^n dx=\frac{a}{n + 1}x^{n+1}+C$ ($n\neq - 1$). $\int(9x^{2}-3x + 4)dx=\int9x^{2}dx-\int3xdx+\int4dx$

Step2: Integrate each term

For $\int9x^{2}dx$, $a = 9$, $n=2$, so $\int9x^{2}dx=9\times\frac{1}{2 + 1}x^{2+1}=3x^{3}$. For $\int3xdx$, $a = 3$, $n = 1$, so $\int3xdx=3\times\frac{1}{1+1}x^{1 + 1}=\frac{3}{2}x^{2}$. For $\int4dx$, $a = 4$, $n=0$, so $\int4dx=4x$.

Step3: Combine results

$\int(9x^{2}-3x + 4)dx=3x^{3}-\frac{3}{2}x^{2}+4x + C$

Answer:

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