Solve f^prime(x)=1/x+x-(1+a)

$$f ^ { \prime } ( x ) = \frac { 1 } { x } + x - ( 1 + a )$$

Solve for a

$$a=x-1+\frac{1}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)xx=1+xx-\left(1+a\right)x$$

Multiply x and x to get x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{2}=1+xx-\left(1+a\right)x$$

Multiply x and x to get x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{2}=1+x^{2}-\left(1+a\right)x$$

Use the distributive property to multiply -1 by 1+a.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{2}=1+x^{2}+\left(-1-a\right)x$$

Use the distributive property to multiply -1-a by x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{2}=1+x^{2}-x-ax$$

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

$$1+x^{2}-x-ax=\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{2}$$

Subtract 1 from both sides.

$$x^{2}-x-ax=\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{2}-1$$

Subtract x^{2} from both sides.

$$-x-ax=\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{2}-1-x^{2}$$

Add x to both sides.

$$-ax=\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{2}-1-x^{2}+x$$

The equation is in standard form.

$$\left(-x\right)a=-x^{2}+x-1$$

Divide both sides by -x.

$$\frac{\left(-x\right)a}{-x}=\frac{-x^{2}+x-1}{-x}$$

Dividing by -x undoes the multiplication by -x.

$$a=\frac{-x^{2}+x-1}{-x}$$

Divide -1-x^{2}+x by -x.

$$a=x-1+\frac{1}{x}$$

Solve for f

$$f\in \mathrm{R}$$

$$a=x-1+\frac{1}{x}\text{ and }x\neq 0$$