Solve ∫ (4x^3-1/x^2) wrt x=x^4+1/x+C

$$\int ( 4 x ^ { 3 } - \frac { 1 } { x ^ { 2 } } ) d x = x ^ { 4 } + \frac { 1 } { x } + C$$

Solve for C

$$C=С$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$x\int 4x^{3}-\frac{1}{x^{2}}\mathrm{d}x=xx^{4}+1+xC$$

To multiply powers of the same base, add their exponents. Add 1 and 4 to get 5.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 4x^{3} times \frac{x^{2}}{x^{2}}.

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

Since \frac{4x^{3}x^{2}}{x^{2}} and \frac{1}{x^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 4x^{3}x^{2}-1.

$$x\int \frac{4x^{5}-1}{x^{2}}\mathrm{d}x=x^{5}+1+xC$$

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

$$x^{5}+1+xC=x\int \frac{4x^{5}-1}{x^{2}}\mathrm{d}x$$

Subtract x^{5} from both sides.

$$1+xC=x\int \frac{4x^{5}-1}{x^{2}}\mathrm{d}x-x^{5}$$

Subtract 1 from both sides.

$$xC=x\int \frac{4x^{5}-1}{x^{2}}\mathrm{d}x-x^{5}-1$$

The equation is in standard form.

$$xC=Сx$$

Divide both sides by x.

$$\frac{xC}{x}=\frac{Сx}{x}$$

Dividing by x undoes the multiplication by x.

$$C=\frac{Сx}{x}$$

Divide Сx by x.

$$C=С$$

Solve for x

$$x\neq 0$$

$$C=С\text{ and }x\neq 0$$