Question

Help me out in this que ....Thank you​

Answer 1

see the attached solution

Answer 2

`(-2x^4-6)/((x+1)(x-1)(x^2+1))`

Explanation:

Given expression:

`(2)/(x+1)-(2x)/(x-1)+(4)/(x^2+1)`

The least common multiplier of the denominators is:

`(x+1)(x-1)(x^2+1)`

Adjust the fractions based on the least common multiplier:

`\implies (2(x-1)(x^2+1))/((x+1)(x-1)(x^2+1))-(2x(x+1)(x^2+1))/((x+1)(x-1)(x^2+1))+(4(x+1)(x-1))/((x+1)(x-1)(x^2+1))`

`\textsf{Apply the fraction rule}: \quad (a)/(c) \pm (b)/(c)=(a \pm b)/(c):`

`\implies (2(x-1)(x^2+1)-2x(x+1)(x^2+1)+4(x+1)(x-1))/((x+1)(x-1)(x^2+1))`

Simplify:

`\implies (2(x^3-x^2+x-1)-2x(x^3+x^2+x+1)+4(x^2-1))/((x+1)(x-1)(x^2+1))`

`\implies (2x^3-2x^2+2x-2-2x^4-2x^3-2x^2-2x+4x^2-4)/((x+1)(x-1)(x^2+1))`

`\implies (-2x^4+2x^3-2x^3-2x^2-2x^2+4x^2+2x-2x-2-4)/((x+1)(x-1)(x^2+1))`

`\implies (-2x^4-6)/((x+1)(x-1)(x^2+1))`