The partial fraction of $\frac{x^2}{x^2+3 x-4}$ is
The partial fraction of $\frac{x^2}{x^2+3 x-4}$ is
- $1+\frac{-16}{5(x+4)}+\frac{1}{5(x-1)}$
- $1+\frac{-1}{x+4}+\frac{1}{x-1}$
- $1+\frac{-13}{5(x+4)}+\frac{1}{5(x-1)}$
- $\frac{2}{x+4}+\frac{1}{x-1}$
Solution
$\because \frac{x^2}{x^2+3 x-4}$
$
\begin{gathered}
\qquad=\frac{x^2}{(x-1)(x+4)}=1+\frac{4-3 x}{(x-1)(x+4)} \\
\text { Let } \frac{-3 x+4}{(x-1)(x+4)}=\frac{A}{x-1}+\frac{B}{x+4} \\
\frac{-3 x+4}{(x-1)(x+4)}=\frac{A(x+4)+B(x-1)}{(x-1)(x+4)} \\
-3 x+4=x(A+B)+(4 A-B)
\end{gathered}
$
On compiaring the coefficients on both sides
$\begin{aligned} & A+B=-3 \text { and } 4 A-B=4 \\ & B=-3-A \\ & \Rightarrow \quad 4 A-(-3-A)=4 \\ & 5 A=1 \\ & A=1 / 5 \\ & \Rightarrow \quad B=-3-\frac{1}{5}=\frac{-16}{5} \\ & \therefore \quad \frac{x^2}{x^2+3 x-4}=1+\frac{1}{5(x-1)}+\frac{-16}{5(x+4)} \\ & \end{aligned}$
Asked in: AP EAMCET 2021 (25 Aug Shift 2)
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