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# Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{x^{3}}{(x-1)(x-2)(x-3)} d x$

Solution:

First we simplify numerator, we will rewrite denominator as shown below

$\frac{x^{3}}{(x-1)(x-2)(x-3)}=\frac{x^{3}}{x^{3}-6 x^{2}+11 x-6}$

Add and subtract numerator with $\left(-6 x^{2}+11 x-6\right)$, we get

$\frac{x^{3}-6 x^{2}+11 x-6+\left(6 x^{2}-11 x+6\right)}{x^{3}-6 x^{2}+11 x-6}$

$\Rightarrow=1+\frac{6 x^{2}-11 x+6}{x^{3}-6 x^{2}+11 x-6}$

$\Rightarrow=1+\frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)}$

The denominator is factorized, so let separate the fraction through partial fraction, hence let

$\frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{x-2}+\frac{C}{x-3}$ ....(i

$\Rightarrow \frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)}$)

$=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}$

$\Rightarrow 6 x^{2}-11 x+6=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) \ldots \ldots$ (ii)

We need to solve for A, B and C. One way to do this is to pick values for $x$ which will cancel each variable.

Put $x=1$ in the above equation, we get

$\Rightarrow 6(1)^{2}-11(1)+6=A(1-2)(1-3)+B(1-1)(1-3)+C(1-1)(1-2)$

$\Rightarrow 1=2 A+0+0$

$\Rightarrow A=\frac{1}{2}$

Now put $x=2$ in equation (ii), we get

$6(2)^{2}-11(2)+6=A(2-2)(2-3)+B(2-1)(2-3)+C(2-1)(2-2)$

$\Rightarrow 8=0-B+0$

$\Rightarrow B=-8$

Now put $x=3$ in equation (ii), we get

$\Rightarrow 6(3)^{2}-11(3)+6=A(3-2)(3-3)+B(3-1)(3-3)+C(3-1)(3-2)$

$\Rightarrow 27=0+0+2 C$

$\Rightarrow C=\frac{27}{2}$

We put the values of $A, B$, and $C$ values back into our partial fractions in equation (i) and replace this as the integrand. We get

$\int\left[1+\frac{6 x^{2}-11 x+6}{(x-1)(x-2)(x-3)}\right] d x$

$\Rightarrow \int\left[1+\frac{\mathrm{A}}{(\mathrm{x}-1)}+\frac{\mathrm{B}}{\mathrm{x}-2}+\frac{\mathrm{C}}{\mathrm{x}-3}\right] \mathrm{dx}$

$\Rightarrow \int\left[1+\frac{\frac{1}{2}}{(\mathrm{x}-1)}+\frac{-8}{\mathrm{x}-2}+\frac{\frac{27}{2}}{\mathrm{x}-3}\right] \mathrm{dx}$

Split up the integral,

$\Rightarrow \int 1 \mathrm{dx}+\frac{1}{2} \int\left[\frac{1}{\mathrm{x}-1}\right] \mathrm{dx}-8 \int\left[\frac{1}{\mathrm{x}-2}\right] \mathrm{dx}+\frac{27}{2} \int\left[\frac{1}{\mathrm{x}-3}\right] \mathrm{dx}$

Let substitute

$u=x-1 \Rightarrow d u=d x$

$y=x-2 \Rightarrow d y=d x$ and

$z=x-3 \Rightarrow d z=d x$, so the above equation becomes,

$\Rightarrow \int 1 \mathrm{dx}+\frac{1}{2} \int\left[\frac{1}{\mathrm{u}}\right] \mathrm{du}-8 \int\left[\frac{1}{\mathrm{y}}\right] \mathrm{dy}+\frac{27}{2} \int\left[\frac{1}{\mathrm{z}}\right] \mathrm{dz}$

On integrating we get

$\Rightarrow x+\frac{1}{2} \log |u|-8 \log |y|+\frac{27}{2} \log |z|+C$

Substituting back, we get

$\Rightarrow x+\frac{1}{2} \log |x-1|-8 \log |x-2|+\frac{27}{2} \log |x-3|+C$

Note: the absolute value signs account for the domain of the natural log function $(x>0)$.

Hence,

$\int \frac{x^{3}}{(x-1)(x-2)(x-3)} d x$

$=x+\frac{1}{2} \log |x-1|-8 \log |x-2|+\frac{27}{2} \log |x-3|+C$

Note: the absolute value signs account for the domain of the natural log function $(x>0)$.

Hence,

$\int \frac{x^{3}}{(x-1)(x-2)(x-3)} d x$

$=x+\frac{1}{2} \log |x-1|-8 \log |x-2|+\frac{27}{2} \log |x-3|+C$