Solve a+1/a-1=a-1/a+1+4a/a^2+1

$$\frac { a + 1 } { a - 1 } = \frac { a - 1 } { a + 1 } + \frac { 4 a } { a ^ { 2 } + 1 }$$

Solve for a

$$a=0$$

Steps for Solving Linear Equation

Variable a cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(a-1\right)\left(a+1\right)\left(a^{2}+1\right), the least common multiple of a-1,a+1,a^{2}+1.

$$\left(a+1\right)\left(a^{2}+1\right)\left(a+1\right)=\left(a-1\right)\left(a^{2}+1\right)\left(a-1\right)+\left(a^{2}-1\right)\times 4a$$

Use the distributive property to multiply a+1 by a^{2}+1.

$$\left(a^{3}+a+a^{2}+1\right)\left(a+1\right)=\left(a-1\right)\left(a^{2}+1\right)\left(a-1\right)+\left(a^{2}-1\right)\times 4a$$

Use the distributive property to multiply a^{3}+a+a^{2}+1 by a+1 and combine like terms.

$$a^{4}+2a^{3}+2a^{2}+2a+1=\left(a-1\right)\left(a^{2}+1\right)\left(a-1\right)+\left(a^{2}-1\right)\times 4a$$

Multiply a-1 and a-1 to get \left(a-1\right)^{2}.

$$a^{4}+2a^{3}+2a^{2}+2a+1=\left(a-1\right)^{2}\left(a^{2}+1\right)+\left(a^{2}-1\right)\times 4a$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-1\right)^{2}.

$$a^{4}+2a^{3}+2a^{2}+2a+1=\left(a^{2}-2a+1\right)\left(a^{2}+1\right)+\left(a^{2}-1\right)\times 4a$$

Use the distributive property to multiply a^{2}-2a+1 by a^{2}+1 and combine like terms.

$$a^{4}+2a^{3}+2a^{2}+2a+1=a^{4}+2a^{2}-2a^{3}-2a+1+\left(a^{2}-1\right)\times 4a$$

Use the distributive property to multiply a^{2}-1 by 4.

$$a^{4}+2a^{3}+2a^{2}+2a+1=a^{4}+2a^{2}-2a^{3}-2a+1+\left(4a^{2}-4\right)a$$

Use the distributive property to multiply 4a^{2}-4 by a.

$$a^{4}+2a^{3}+2a^{2}+2a+1=a^{4}+2a^{2}-2a^{3}-2a+1+4a^{3}-4a$$

Combine -2a^{3} and 4a^{3} to get 2a^{3}.

$$a^{4}+2a^{3}+2a^{2}+2a+1=a^{4}+2a^{2}+2a^{3}-2a+1-4a$$

Combine -2a and -4a to get -6a.

$$a^{4}+2a^{3}+2a^{2}+2a+1=a^{4}+2a^{2}+2a^{3}-6a+1$$

Subtract a^{4} from both sides.

$$a^{4}+2a^{3}+2a^{2}+2a+1-a^{4}=2a^{2}+2a^{3}-6a+1$$

Combine a^{4} and -a^{4} to get 0.

$$2a^{3}+2a^{2}+2a+1=2a^{2}+2a^{3}-6a+1$$

Subtract 2a^{2} from both sides.

$$2a^{3}+2a^{2}+2a+1-2a^{2}=2a^{3}-6a+1$$

Combine 2a^{2} and -2a^{2} to get 0.

$$2a^{3}+2a+1=2a^{3}-6a+1$$

Subtract 2a^{3} from both sides.

$$2a^{3}+2a+1-2a^{3}=-6a+1$$

Combine 2a^{3} and -2a^{3} to get 0.

$$2a+1=-6a+1$$

Add 6a to both sides.

$$2a+1+6a=1$$

Combine 2a and 6a to get 8a.

$$8a+1=1$$

Subtract 1 from both sides.

$$8a=1-1$$

Subtract 1 from 1 to get 0.

$$8a=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since 8 is not equal to 0, a must be equal to 0.

$$a=0$$