$$int 4 x ^ { 6 } - 2 x ^ { 3 } + 7 x - 4 d x$$

Explanation:

Step1: Integrate each term separately

$$\int 4x^{6}dx-\int 2x^{3}dx+\int 7xdx-\int 4dx$$

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

• For $\int 4x^{6}dx$: $4\times\frac{x^{6 + 1}}{6+1}=\frac{4x^{7}}{7}$
• For $\int 2x^{3}dx$: $2\times\frac{x^{3+1}}{3 + 1}=\frac{2x^{4}}{4}=\frac{x^{4}}{2}$
• For $\int 7xdx$: $7\times\frac{x^{1+1}}{1+1}=\frac{7x^{2}}{2}$
• For $\int 4dx$: $4x$

Step3: Combine the results

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

Answer:

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