Evaluate the following integral:
Question:

Evaluate the following integral:

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

Solution:

Denominator is already factorized, so let

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

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

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

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

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

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

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

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

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

$\Rightarrow 9=0+0+2 C$

$\Rightarrow C=\frac{9}{2}$

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

$\int\left[\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\right] d x$

$\Rightarrow \int\left[\frac{\frac{1}{2}}{x-1}+\frac{-4}{x-2}+\frac{\frac{9}{2}}{x-3}\right] d x$

Split up the integral,

$\Rightarrow \frac{1}{2} \int\left[\frac{1}{x-1}\right] d x-4 \int\left[\frac{1}{x-2}\right] d x+\frac{9}{2} \int\left[\frac{1}{x-3}\right] d x$

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,

$\Rightarrow \frac{1}{2} \int\left[\frac{1}{\mathrm{u}}\right] \mathrm{du}-4 \int\left[\frac{1}{\mathrm{y}}\right] \mathrm{dy}+\frac{9}{2} \int\left[\frac{1}{\mathrm{z}}\right] \mathrm{dz}$

On integrating we get

$\Rightarrow \frac{1}{2} \log |u|-4 \log |y|+\frac{9}{2} \log |z|+C$

Substituting back, we get

$\Rightarrow \frac{1}{2} \log |x-1|-4 \log |x-2|+\frac{9}{2} \log |x-3|+C$

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

Hence,

$\int \frac{x^{2}}{(x-1)(x-2)(x-3)} d x=\frac{1}{2} \log |x-1|-4 \log |x-2|+\frac{9}{2} \log |x-3|+C$

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