Solve ∫ x wrt x=x^4-1/a+1

$$\int x d x = \frac { x ^ { 4 - 1 } } { a + 1 }$$

Solve for a

$$\left\{\begin{matrix}a=\frac{2x^{3}-x^{2}-С}{x^{2}+С_{1}}\text{, }&С\neq -x^{2}\text{ and }x\neq 0\text{ and }\left(С_{1}<0\text{ or }|x|\neq \frac{\sqrt{С_{2}}}{2}\right)\text{ and }С_{3}\neq -\frac{x^{2}}{2}\\a\neq -1\text{, }&С=0\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable a cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by a+1.

$$\left(a+1\right)\int x\mathrm{d}x=x^{4-1}$$

Use the distributive property to multiply a+1 by \int x\mathrm{d}x.

$$a\int x\mathrm{d}x+\int x\mathrm{d}x=x^{4-1}$$

Subtract 1 from 4 to get 3.

$$a\int x\mathrm{d}x+\int x\mathrm{d}x=x^{3}$$

Subtract \int x\mathrm{d}x from both sides.

$$a\int x\mathrm{d}x=x^{3}-\int x\mathrm{d}x$$

The equation is in standard form.

$$\left(\frac{x^{2}}{2}+С\right)a=x^{3}-\frac{x^{2}}{2}-С$$

Divide both sides by \frac{1}{2}x^{2}+С.

$$\frac{\left(\frac{x^{2}}{2}+С\right)a}{\frac{x^{2}}{2}+С}=\frac{x^{3}-\frac{x^{2}}{2}-С}{\frac{x^{2}}{2}+С}$$

Dividing by \frac{1}{2}x^{2}+С undoes the multiplication by \frac{1}{2}x^{2}+С.

$$a=\frac{x^{3}-\frac{x^{2}}{2}-С}{\frac{x^{2}}{2}+С}$$

Divide x^{3}-\frac{x^{2}}{2}-С by \frac{1}{2}x^{2}+С.

$$a=\frac{2x^{3}-x^{2}-2С}{x^{2}+2С_{1}}$$

Variable a cannot be equal to -1.

$$a=\frac{2x^{3}-x^{2}-2С}{x^{2}+2С_{1}}\text{, }a\neq -1$$