Evalute : 2 x+1 x(x-1)(x-4) d x - Vidyadip

Question

Evalute : $\int \frac{2 x+1}{x(x-1)(x-4)} d x$

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Answer

Let $I =\int \frac{2 x+1}{x(x-1)(x-4)} d x$
Let $\int \frac{2 x+1}{x(x-1)(x-4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-4}$
$
\therefore 2 x +1= A ( x -1)( x -4)+ Bx ( x -4)+ Cx ( x -1)
$
Put $x =0$, we get
$
\begin{aligned}
& 2(0)+1= A (-1)(-4)+ B (0)(-4)+ C (0)(-1) \\
& \therefore 1=4 A \\
& \therefore A =\frac{1}{4}
\end{aligned}
$
Put $x-1=0$, i.e. $x=1$, we get
$
\begin{aligned}
& 2(1)+1= A (0)(-3)+ B (1)(-3)+ C (1)(0) \\
& \therefore 3=-3 B \\
& \therefore B =-1
\end{aligned}
$
Put $x-4=0$, i.e $x=4$, we get
$
\begin{aligned}
& 2(4)+1= A (3)(0)+ B (4)(0)+ C (4)(3) \\
& \therefore 9=12 C \\
& \therefore C =\frac{3}{4}
\end{aligned}
$
$
\therefore \frac{2 x+1}{x(x-1)(x-4)}=\frac{\left(\frac{1}{4}\right)}{x}+\frac{(-1)}{x-1}+\frac{\left(\frac{3}{4}\right)}{x-4}
$
$
\begin{aligned}
\therefore I & =\int\left[\frac{\left(\frac{1}{4}\right)}{x}+\frac{(-1)}{x-1}+\frac{\left(\frac{3}{4}\right)}{x-4}\right] d x \\
& =\frac{1}{4} \int \frac{1}{x} d x-\int \frac{1}{x-1} d x+\frac{3}{4} \int \frac{1}{x-4} d x \\
& =\frac{1}{4} \log |x|-\log |x-1|+\frac{3}{4} \log |x-4|+c .
\end{aligned}
$

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