Solve ( 2 x - 1 / x ^ 2 ) wrt x = ∫ (from 1 to 3) of 2 x d x - int_1^31/x^2dx

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

Solve for d

$$d=\frac{22x}{3\left(2x^{3}-1\right)}$$

$$x\neq \frac{2^{\frac{2}{3}}}{2}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x^{2}.

$$\left(2x-\frac{1}{x^{2}}\right)dxx^{2}=x^{2}\int _{1}^{3}2x\mathrm{d}x-x^{2}\int _{1}^{3}\frac{1}{x^{2}}\mathrm{d}x$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\left(2x-\frac{1}{x^{2}}\right)dx^{3}=x^{2}\int _{1}^{3}2x\mathrm{d}x-x^{2}\int _{1}^{3}\frac{1}{x^{2}}\mathrm{d}x$$

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

$$\left(\frac{2xx^{2}}{x^{2}}-\frac{1}{x^{2}}\right)dx^{3}=x^{2}\int _{1}^{3}2x\mathrm{d}x-x^{2}\int _{1}^{3}\frac{1}{x^{2}}\mathrm{d}x$$

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

$$\frac{2xx^{2}-1}{x^{2}}dx^{3}=x^{2}\int _{1}^{3}2x\mathrm{d}x-x^{2}\int _{1}^{3}\frac{1}{x^{2}}\mathrm{d}x$$

Do the multiplications in 2xx^{2}-1.

$$\frac{2x^{3}-1}{x^{2}}dx^{3}=x^{2}\int _{1}^{3}2x\mathrm{d}x-x^{2}\int _{1}^{3}\frac{1}{x^{2}}\mathrm{d}x$$

Express \frac{2x^{3}-1}{x^{2}}d as a single fraction.

$$\frac{\left(2x^{3}-1\right)d}{x^{2}}x^{3}=x^{2}\int _{1}^{3}2x\mathrm{d}x-x^{2}\int _{1}^{3}\frac{1}{x^{2}}\mathrm{d}x$$

Express \frac{\left(2x^{3}-1\right)d}{x^{2}}x^{3} as a single fraction.

$$\frac{\left(2x^{3}-1\right)dx^{3}}{x^{2}}=x^{2}\int _{1}^{3}2x\mathrm{d}x-x^{2}\int _{1}^{3}\frac{1}{x^{2}}\mathrm{d}x$$

Cancel out x^{2} in both numerator and denominator.

$$dx\left(2x^{3}-1\right)=x^{2}\int _{1}^{3}2x\mathrm{d}x-x^{2}\int _{1}^{3}\frac{1}{x^{2}}\mathrm{d}x$$

Use the distributive property to multiply dx by 2x^{3}-1.

$$2dx^{4}-dx=x^{2}\int _{1}^{3}2x\mathrm{d}x-x^{2}\int _{1}^{3}\frac{1}{x^{2}}\mathrm{d}x$$

Combine all terms containing d.

$$\left(2x^{4}-x\right)d=x^{2}\int _{1}^{3}2x\mathrm{d}x-x^{2}\int _{1}^{3}\frac{1}{x^{2}}\mathrm{d}x$$

The equation is in standard form.

$$\left(2x^{4}-x\right)d=\frac{22x^{2}}{3}$$

Divide both sides by 2x^{4}-x.

$$\frac{\left(2x^{4}-x\right)d}{2x^{4}-x}=\frac{22x^{2}}{3\left(2x^{4}-x\right)}$$

Dividing by 2x^{4}-x undoes the multiplication by 2x^{4}-x.

$$d=\frac{22x^{2}}{3\left(2x^{4}-x\right)}$$

Divide \frac{22x^{2}}{3} by 2x^{4}-x.

$$d=\frac{22x}{3\left(2x^{3}-1\right)}$$