Question:
Evaluate the integral:
\[
\int \frac{x}{x^2-5x+4}\,dx
\]
Evaluate the integral:
\[
\int \frac{x}{x^2-5x+4}\,dx
\]
Show Hint
Whenever the denominator factorizes into distinct linear factors, partial fractions is usually the fastest integration technique.
Updated On: Jun 17, 2026
- \(\displaystyle \frac{1}{3}\log\left|\frac{(x-4)^4}{x-1}\right|+c\)
- \(\displaystyle \frac{4}{3}\log\left|\frac{x-4}{(x-1)^4}\right|+c\)
- \(\displaystyle \frac{1}{3}\log\left|\frac{(x-4)^2}{x-1}\right|+c\)
- \(\displaystyle \frac{4}{3}\log\left|\frac{x-4}{(x-1)^4}\right|+c\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
To evaluate rational integrals, we factorize the denominator and then apply partial fractions.
\[
x^2-5x+4=(x-1)(x-4)
\]
Hence,
\[
\frac{x}{x^2-5x+4}=\frac{x}{(x-1)(x-4)}
\]
We now decompose into partial fractions.
Step 1: Resolve into partial fractions. Assume \[ \frac{x}{(x-1)(x-4)}=\frac{A}{x-1}+\frac{B}{x-4} \] Multiplying throughout by \((x-1)(x-4)\), \[ x=A(x-4)+B(x-1) \] \[ x=(A+B)x-(4A+B) \] Comparing coefficients, \[ A+B=1 \] and \[ 4A+B=0 \] From \(B=-4A\), \[ A-4A=1 \] \[ -3A=1 \] \[ A=-\frac13 \] Then, \[ B=\frac43 \] Therefore, \[ \frac{x}{(x-1)(x-4)} =-\frac1{3(x-1)}+\frac4{3(x-4)} \]
Step 2: Integrate term by term. \[ \int \frac{x}{x^2-5x+4}\,dx = -\frac13\int\frac{dx}{x-1} +\frac43\int\frac{dx}{x-4} \] Using \[ \int \frac{dx}{x-a}=\log|x-a| \] we get \[ = -\frac13\log|x-1| +\frac43\log|x-4| +c \] Taking \(\frac13\) common, \[ = \frac13\left[4\log|x-4|-\log|x-1|\right]+c \] \[ = \frac13\log\left|\frac{(x-4)^4}{x-1}\right|+c \] Hence the required answer is \[ \boxed{ \frac13\log\left|\frac{(x-4)^4}{x-1}\right|+c } \]
Step 1: Resolve into partial fractions. Assume \[ \frac{x}{(x-1)(x-4)}=\frac{A}{x-1}+\frac{B}{x-4} \] Multiplying throughout by \((x-1)(x-4)\), \[ x=A(x-4)+B(x-1) \] \[ x=(A+B)x-(4A+B) \] Comparing coefficients, \[ A+B=1 \] and \[ 4A+B=0 \] From \(B=-4A\), \[ A-4A=1 \] \[ -3A=1 \] \[ A=-\frac13 \] Then, \[ B=\frac43 \] Therefore, \[ \frac{x}{(x-1)(x-4)} =-\frac1{3(x-1)}+\frac4{3(x-4)} \]
Step 2: Integrate term by term. \[ \int \frac{x}{x^2-5x+4}\,dx = -\frac13\int\frac{dx}{x-1} +\frac43\int\frac{dx}{x-4} \] Using \[ \int \frac{dx}{x-a}=\log|x-a| \] we get \[ = -\frac13\log|x-1| +\frac43\log|x-4| +c \] Taking \(\frac13\) common, \[ = \frac13\left[4\log|x-4|-\log|x-1|\right]+c \] \[ = \frac13\log\left|\frac{(x-4)^4}{x-1}\right|+c \] Hence the required answer is \[ \boxed{ \frac13\log\left|\frac{(x-4)^4}{x-1}\right|+c } \]
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