Evaluate the following integral:
$\int \frac{5 x^{2}-1}{x(x-1)(x+1)} d x$
Denominator is factorized, so let separate the fraction through partial fraction, hence let
$\frac{5 x^{2}-1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1} \ldots \ldots$ (i)
$\Rightarrow \frac{5 x^{2}-1}{x(x-1)(x+1)}=\frac{A(x-1)(x+1)+B x(x+1)+C x(x-1)}{x(x-1)(x+1)}$
$\Rightarrow 5 x^{2}-1=A(x-1)(x+1)+B x(x+1)+C x(x-1) \ldots \ldots .($ 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 $x=0$ in the above equation, we get
$\Rightarrow 5(0)^{2}-1=A(0-1)(0+1)+B(0)(0+1)+C(0)(0-1)$
$\Rightarrow A=1$
Now put $x=1$ in equation (ii), we get
$\Rightarrow 5(1)^{2}-1=A(1-1)(1+1)+B(1)(1+1)+C(1)(1-1)$
$\Rightarrow 4=0+2 B+0$
$\Rightarrow B=2$
Now put $x=-1$ in equation (ii), we get
$\Rightarrow 5(-1)^{2}-1=A(-1-1)(-1+1)+B(-1)(-1+1)+C(-1)(-1-1)$
$\Rightarrow 4=0+0+2 C$
$\Rightarrow C=2$
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
$\int\left[\frac{5 x^{2}-1}{x(x-1)(x+1)}\right] d x$
$\Rightarrow \int\left[\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\right] d x$
$\Rightarrow \int\left[\frac{1}{x}+\frac{2}{x-1}+\frac{2}{x+1}\right] d x$
Split up the integral,
$\Rightarrow \int\left[\frac{1}{x}\right] d x+2 \int\left[\frac{1}{x-1}\right] d x+2 \int\left[\frac{1}{x+1}\right] d x$
Let substitute
$u=x-1 \Rightarrow d u=d x$
$y=x+1 \Rightarrow d y=d x$, so the above equation becomes,
$\Rightarrow \int\left[\frac{1}{x}\right] d x+2 \int\left[\frac{1}{u}\right] d u+2 \int\left[\frac{1}{y}\right] d y$
On integrating we get
$\Rightarrow \log |x|+2 \log |u|+2 \log |y|+C$
Substituting back, we get
$\Rightarrow \log |x|+2 \log |x-1|+2 \log |x+1|+C$
Applying logarithm rule, we get
$\Rightarrow \log |x|+\log \left|(x-1)^{2}\right|+\log \left|(x+1)^{2}\right|+C$
$\Rightarrow \log \left|x\left(x^{2}-1\right)^{2}\right|+C$
Note: the absolute value signs account for the domain of the natural log function $(x>0)$.
Hence,
$\int \frac{5 x^{2}-1}{x(x-1)(x+1)} d x=\log \left|x\left(x^{2}-1\right)^{2}\right|+C$
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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,