Solve ∫ x^4/1+x^2 wrt x=x^3/3-x+arctanx+C

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

Solve for C

$$C=С$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 3.

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

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

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

Subtract x^{3} from both sides.

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

Add 3x to both sides.

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

Subtract 3\arctan(x) from both sides.

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

The equation is in standard form.

$$3C=3\left(\arctan(x)-\arctan(x)+С\right)$$

Divide both sides by 3.

$$\frac{3C}{3}=\frac{С}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$C=\frac{С}{3}$$

Solve for x

$$x\in \mathrm{R}$$

$$C=С$$