∫3x−1(x−1)(x−2)(x−3)dx is equal to

# $\int \frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}\mathrm{dx}$ is equal to

1. A

$\mathrm{log}|\mathrm{x}-1|-5|\mathrm{log}|\mathrm{x}-2|+4\mathrm{log}|\mathrm{x}-3\mid +\mathrm{C}$

2. B

$|\mathrm{log}|\mathrm{x}-1|-\mathrm{log}|\mathrm{x}-2|+4|\mathrm{log}|\mathrm{x}-3|+\mathrm{C}$

3. C

$5\mathrm{log}|\mathrm{x}-1|-\mathrm{log}|\mathrm{x}-2|+4\mathrm{log}|\mathrm{x}-3|+\mathrm{C}$

4. D

None of the above

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### Solution:

Let $\frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}=\frac{\mathrm{A}}{\left(\mathrm{x}-1\right)}+\frac{\mathrm{B}}{\left(\mathrm{x}-2\right)}+\frac{\mathrm{C}}{\left(\mathrm{x}-3\right)}$
$⇒\frac{3\mathrm{x}-1}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}=\frac{\mathrm{A}\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)+\mathrm{B}\left(\mathrm{x}-1\right)\left(\mathrm{x}-3\right)+\mathrm{C}\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)}{\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)}$
$\begin{array}{l}⇒3\mathrm{x}-1=\mathrm{A}\left[{\mathrm{x}}^{2}-5\mathrm{x}+6\right]+\mathrm{B}\left[{\mathrm{x}}^{2}-4\mathrm{x}+3\right]+\mathrm{C}\left[{\mathrm{x}}^{2}-3\mathrm{x}+2\right]\\ ⇒3\mathrm{x}-1={\mathrm{x}}^{2}\left(\mathrm{A}+\mathrm{B}+\mathrm{C}\right)+\mathrm{x}\left(-5\mathrm{A}-4\mathrm{B}-3\mathrm{C}\right)+\left(6\mathrm{A}+3\mathrm{B}+2\mathrm{C}\right)\end{array}$
On equating the coefficients of x2 ,x and constant term on both sides, we get

From Eq. (i), we get A = -(B + C)
On putting the value of A in Eqs. (ii) and (iii), we get
$-5\left\{-\left(\mathrm{B}+\mathrm{C}\right)\right\}-4\mathrm{B}-3\mathrm{C}=3$

On solving Eqs. (iv) and (v), we get C = 4 On putting the value of C in Eq. (iv), we get
$\mathrm{B}+2×4=3$

On putting the value of Band C in Eq. (i), we get

Now,

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