Evaluate the following integral:
$\int \frac{\mathrm{x}^{2}}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)} \mathrm{dx}$
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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All Study Material
- JEE Main
- Exam Pattern
- Previous Year Papers
- PYQ Chapterwise
- Physics
- Kinematics 1D
- Kinemetics 2D
- Friction
- Work, Power, Energy
- Centre of Mass and Collision
- Rotational Dynamics
- Gravitation
- Calorimetry
- Elasticity
- Thermal Expansion
- Heat Transfer
- Kinetic Theory of Gases
- Thermodynamics
- Simple Harmonic Motion
- Wave on String
- Sound waves
- Fluid Mechanics
- Electrostatics
- Current Electricity
- Capacitor
- Magnetism and Matter
- Electromagnetic Induction
- Atomic Structure
- Dual Nature of Matter
- Nuclear Physics
- Radioactivity
- Semiconductors
- Communication System
- Error in Measurement & instruments
- Alternating Current
- Electromagnetic Waves
- Wave Optics
- X-Rays
- All Subjects
- Physics
- Motion in a Plane
- Law of Motion
- Work, Energy and Power
- Systems of Particles and Rotational Motion
- Gravitation
- Mechanical Properties of Solids
- Mechanical Properties of Fluids
- Thermal Properties of matter
- Thermodynamics
- Kinetic Theory
- Oscillations
- Waves
- Electric Charge and Fields
- Electrostatic Potential and Capacitance
- Current Electricity
- Thermoelectric Effects of Electric Current
- Heating Effects of Electric Current
- Moving Charges and Magnetism
- Magnetism and Matter
- Electromagnetic Induction
- Alternating Current
- Electromagnetic Wave
- Ray Optics and Optical Instruments
- Wave Optics
- Dual Nature of Radiation and Matter
- Atoms
- Nuclei
- Semiconductor Electronics: Materials, Devices and Simple Circuits.
- Chemical Effects of Electric Current,