Solve ∫ (7x-2)d=(7x-2)^4/28+c

$$\int ( 7 x - 2 ) d = \frac { ( 7 x - 2 ) ^ { 4 } } { 28 } + c$$

Solve for c

$$c=-\frac{\left(7x-2\right)^{4}}{28}+7dx^{2}+Сx-2dx$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 28.

$$28\int 7x-2\mathrm{d}dx=\left(7x-2\right)^{4}+28c$$

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

$$\left(7x-2\right)^{4}+28c=28\int 7x-2\mathrm{d}dx$$

Subtract \left(7x-2\right)^{4} from both sides.

$$28c=28\int 7x-2\mathrm{d}dx-\left(7x-2\right)^{4}$$

The equation is in standard form.

$$28c=28x\left(d\left(7x-2\right)+С\right)-\left(7x-2\right)^{4}$$

Divide both sides by 28.

$$\frac{28c}{28}=\frac{28x\left(d\left(7x-2\right)+С\right)-\left(7x-2\right)^{4}}{28}$$

Dividing by 28 undoes the multiplication by 28.

$$c=\frac{28x\left(d\left(7x-2\right)+С\right)-\left(7x-2\right)^{4}}{28}$$

Divide 28\left(\left(7x-2\right)d+С\right)x-\left(7x-2\right)^{4} by 28.

$$c=x\left(d\left(7x-2\right)+С\right)-\frac{\left(7x-2\right)^{4}}{28}$$