Solve ∫ x(z+x^7) wrt x=(z+x)^2/6+c

$$\int x ( z + x ^ { 7 } ) d x = \frac { ( z + x ) ^ { 2 } } { 6 } + c$$

Solve for c

$$c=\frac{x^{9}}{9}+\frac{zx^{2}}{2}-\frac{\left(x+z\right)^{2}}{6}+С$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 6.

$$6\int x\left(z+x^{7}\right)\mathrm{d}x=\left(z+x\right)^{2}+6c$$

Use the distributive property to multiply x by z+x^{7}.

$$6\int xz+x^{8}\mathrm{d}x=\left(z+x\right)^{2}+6c$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(z+x\right)^{2}.

$$6\int xz+x^{8}\mathrm{d}x=z^{2}+2zx+x^{2}+6c$$

Swap sides so that all variable terms are on the left hand side.

$$z^{2}+2zx+x^{2}+6c=6\int xz+x^{8}\mathrm{d}x$$

Subtract z^{2} from both sides.

$$2zx+x^{2}+6c=6\int xz+x^{8}\mathrm{d}x-z^{2}$$

Subtract 2zx from both sides.

$$x^{2}+6c=6\int xz+x^{8}\mathrm{d}x-z^{2}-2zx$$

Subtract x^{2} from both sides.

$$6c=6\int xz+x^{8}\mathrm{d}x-z^{2}-2zx-x^{2}$$

The equation is in standard form.

$$6c=\frac{2x^{9}}{3}+3zx^{2}-2xz-x^{2}-z^{2}+6С$$

Divide both sides by 6.

$$\frac{6c}{6}=\frac{\frac{2x^{9}}{3}+3zx^{2}-\left(x+z\right)^{2}+6С}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$c=\frac{\frac{2x^{9}}{3}+3zx^{2}-\left(x+z\right)^{2}+6С}{6}$$

Divide 3zx^{2}+\frac{2x^{9}}{3}+6С-\left(z+x\right)^{2} by 6.

$$c=\frac{x^{9}}{9}+\frac{zx^{2}}{2}-\frac{\left(x+z\right)^{2}}{6}+С$$