find $int (11x^{3}+5x^{2}-5x + 3)mathrm{d}x$.\n$int (11x^{3}+5x^{2}-5x + 3)mathrm{d}x=square$

Explanation:

Step1: Apply sum - rule of integration

$\int(11x^{3}+5x^{2}-5x + 3)dx=\int11x^{3}dx+\int5x^{2}dx-\int5xdx+\int3dx$

Step2: Use power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$)

For $\int11x^{3}dx$, since $\int x^{3}dx=\frac{x^{4}}{4}$, then $\int11x^{3}dx = 11\times\frac{x^{4}}{4}=\frac{11x^{4}}{4}$; for $\int5x^{2}dx$, since $\int x^{2}dx=\frac{x^{3}}{3}$, then $\int5x^{2}dx = 5\times\frac{x^{3}}{3}=\frac{5x^{3}}{3}$; for $\int5xdx$, since $\int xdx=\frac{x^{2}}{2}$, then $\int5xdx = 5\times\frac{x^{2}}{2}=\frac{5x^{2}}{2}$; for $\int3dx$, since $\int dx=x$, then $\int3dx = 3x$.

Step3: Combine the results

$\int(11x^{3}+5x^{2}-5x + 3)dx=\frac{11x^{4}}{4}+\frac{5x^{3}}{3}-\frac{5x^{2}}{2}+3x + C$

Answer:

$\frac{11x^{4}}{4}+\frac{5x^{3}}{3}-\frac{5x^{2}}{2}+3x + C$