The partial fraction of {x²}{x²+3x-4} is ______

Correct Answer is : 1+−165(x+4)+15(x−1)1+-16/5(x+4)+1/5(x-1)1+5(x+4)−16+5(x−1)1

Correct Answer is : 1+−165(x+4)+15(x−1)1+-16/5(x+4)+1/5(x-1)1+5(x+4)−16+5(x−1)1

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AP EAPCET 25-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 19 of 160

Marks: +1, -0

The partial fraction of

\frac{x^{2}}{x^{2}+3x-4}

is ______

Solution: 👈: Video Solution

∵

\frac{x^{2}}{x^{2}+3x-4}

=\frac{x^{2}}{(x-1)(x+4)}=1+\frac{4-3x}{(x-1)(x+4)}

Let

\frac{-3x+4}{(x-1)(x+4)}=\frac{A}{x-1}+\frac{B}{x+4}

\frac{-3x+4}{(x-1)(x+4)}=\frac{A(x+4)+B(x-1)}{(x-1)(x+4)}

-3x+4=x(A+B)+(4A-B)

On compiaring the coefficients on both sides

A+B=-3

and

4A-B=4

B=-3-A

⇒

4A-(-3-A)=4

5A=1

A=\frac{1}{5}

⇒

B=-3-\frac{1}{5}=\frac{-16}{5}

∴

\frac{x^{2}}{x^{2}+3x-4}=1+\frac{1}{5(x-1)}+\frac{-16}{5(x+4)}

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