Answer
\[I=x+\frac{6}{5}\ln |x-1|-\frac{21}{5}\ln |x+4|+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int\frac{x^2+5}{(x-1)(x+4)}dx\]
\[I=\int\frac{x^2+5}{x^2+3x-4}dx\]
\[I=\int\frac{(x^2+3x-4)+(-3x+9)}{x^2+3x-4}dx\]
\[I=\int\left[1-\frac{3(x-3)}{(x-1)(x+4)}\right]dx\]
\[I=\int dx-3\int\frac{(x-3)}{(x-1)(x+4)}dx\]
We first determine the partial fraction decomposition of $\frac{(x-3)}{(x-1)(x+4)}$
\[\frac{(x-3)}{(x-1)(x+4)}=\frac{A}{x-1}+\frac{B}{x+4}\]
\[x-3=A(x+4)+B(x-1)\]
\[x-3=(A+B)x+(4A-B)\]
Comparing both sides
\[A+B=1\;\;\;...(1)\]
\[4A-B=-3\;\;\;...(2)\]
Add (1) and (2)
\[5A=-2\Rightarrow A=\frac{-2}{5}\]
From (1)
\[B=\frac{7}{5}\]
\[\frac{(x-3)}{(x-1)(x+4)}=\frac{-2}{5(x-1)}+\frac{7}{5(x+4)}\]
\[\Rightarrow I=\int dx-3\int\left[\frac{-2}{5(x-1)}+\frac{7}{5(x+4)}\right]dx\]
\[I=\int dx+\frac{6}{5}\int\frac{1}{(x-1)}dx-\frac{21}{5}\int\frac{1}{(x+4)}dx\]
\[I=x+\frac{6}{5}\ln |x-1|-\frac{21}{5}\ln |x+4|+C\]
Where $C$ is constant of integration
Hence,\[I= x+\frac{6}{5}\ln |x-1|-\frac{21}{5}\ln |x+4|+C.\]
