Calculate: {\text{begin}array l 3x+2y=6 } 5x+4y=8\text{end}array .

Expression: $$\int ( 5 x ^ { 4 } - 4 x ^ { 3 } + 3 x ^ { 2 } - 6 x + 1 ) d x$$

Integrate the sum term by term.

$$\int 5x^{4}\mathrm{d}x+\int -4x^{3}\mathrm{d}x+\int 3x^{2}\mathrm{d}x+\int -6x\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$5\int x^{4}\mathrm{d}x-4\int x^{3}\mathrm{d}x+3\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 1\mathrm{d}x$$

Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{4}\mathrm{d}x$ with $\frac{x^{5}}{5}$. Multiply $5$ times $\frac{x^{5}}{5}$.

$$x^{5}-4\int x^{3}\mathrm{d}x+3\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 1\mathrm{d}x$$

Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{3}\mathrm{d}x$ with $\frac{x^{4}}{4}$. Multiply $-4$ times $\frac{x^{4}}{4}$.

$$x^{5}-x^{4}+3\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 1\mathrm{d}x$$

Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{2}\mathrm{d}x$ with $\frac{x^{3}}{3}$. Multiply $3$ times $\frac{x^{3}}{3}$.

$$x^{5}-x^{4}+x^{3}-6\int x\mathrm{d}x+\int 1\mathrm{d}x$$

Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x\mathrm{d}x$ with $\frac{x^{2}}{2}$. Multiply $-6$ times $\frac{x^{2}}{2}$.

$$x^{5}-x^{4}+x^{3}-3x^{2}+\int 1\mathrm{d}x$$

Find the integral of $1$ using the table of common integrals rule $\int a\mathrm{d}x=ax$.

$$x^{5}-x^{4}+x^{3}-3x^{2}+x$$

If $F\left(x\right)$ is an antiderivative of $f\left(x\right)$, then the set of all antiderivatives of $f\left(x\right)$ is given by $F\left(x\right)+C$. Therefore, add the constant of integration $C\in \mathrm{R}$ to the result.

$$x^{5}-x^{4}+x^{3}-3x^{2}+x+С$$