Rational Integrals and Partial Fractions: Original Exercises
Exercise editors and source: Prof. Antonino De Martino and Dr. Luana Manfredini, Eserciziario 2.1. The original exercises and source theory are presented here in English for Logic & Motion.
Polynomial Division and Hermite Decomposition
These eight original exercises cover proper and improper rational functions, distinct linear factors, irreducible quadratic factors and repeated poles. Factor the denominator, divide first if needed, and match numerator coefficients. Primitives are valid separately on each interval avoiding the poles.
Exercise 1
Evaluate the integral:
\[\int \frac{1}{x^{3}-1} \, dx.\]Solution.
Factor the denominator and use the source’s Hermite decomposition: \(x^3-1=(x-1)(x^2+x+1).\) \(\frac1{x^3-1}=\frac A{x-1}+\frac{Bx+C}{x^2+x+1}.\) Matching the numerator coefficients gives: \(A+B=0,\quad A-B+C=0,\quad A-C=1.\) \(A=\frac13,\quad B=-\frac13,\quad C=-\frac23.\) \(I=\frac13\int\frac{dx}{x-1}-\frac13\int\frac{x+2}{x^2+x+1}\,dx.\) Separate a logarithmic derivative: \(\frac13\int\frac{x+2}{x^2+x+1}\,dx =\frac16\int\frac{2x+1}{x^2+x+1}\,dx +\frac12\int\frac{dx}{x^2+x+1}.\) Complete the square, or use the complex conjugate factors as in the source: \(x^2+x+1=(x+1/2)^2+3/4 =(x+1/2-i\sqrt3/2)(x+1/2+i\sqrt3/2).\) \(\frac12\int\frac{dx}{x^2+x+1} =\frac1{\sqrt3}\arctan\left(\frac{2x+1}{\sqrt3}\right)+C.\)
Final result
\[\frac13\log|x-1|-\frac16\log(x^2+x+1)-\frac1{\sqrt3}\arctan\left(\frac{2x+1}{\sqrt3}\right)+C\]Exercise 2
Evaluate the integral:
\[\int \frac{x^{3}}{x^{2}-5x+6}\, dx.\]Solution.
Start with polynomial division: \(\frac{x^3}{x^2-5x+6}=x+5+\frac{19x-30}{(x-2)(x-3)}.\) The first two terms integrate to x²/2 + 5x. For the remainder: \(\frac{19x-30}{(x-2)(x-3)}=\frac A{x-2}+\frac B{x-3}.\) \(A+B=19,\quad3A+2B=30, \qquad A=-8,\quad B=27.\) \(\int\frac{19x-30}{(x-2)(x-3)}\,dx =-8\log|x-2|+27\log|x-3|+C.\)
Final result
\[\frac{x^2}{2}+5x-8\log|x-2|+27\log|x-3|+C\]Exercise 3
Evaluate the integral:
\[\int \frac{x^{2}+1}{x^{3}-4x^{2}+5x-2}\, dx.\]Solution.
The denominator factors as: \(x^3-4x^2+5x-2=(x-1)^2(x-2).\) Author’s observation: one strategy is to find the most immediate root of the cubic and then solve the remaining quadratic equation. The repeated factor requires a second-order term. Use the source’s derivative form: \(\frac{x^2+1}{(x-1)^2(x-2)} =\frac A{x-1}+\frac B{x-2}+\frac d{dx}\left(\frac C{x-1}\right).\) The combined numerator is: \((A+B)x^2-(3A+2B+C)x+(2A+B+2C).\) \(A+B=1,\quad3A+2B+C=0,\quad2A+B+2C=1.\) \(A=-4,\quad B=5,\quad C=2.\) Integrate the logarithmic terms and the exact derivative separately.
Final result
\[-4\log|x-1|+5\log|x-2|+\frac2{x-1}+C\]Exercise 4
Solve the following exercise:
\[\int \frac{\tan^{3} x + \tan x}{\tan x+4} \, dx\]Solution.
Set t=tan x and dt=(1+tan²x)dx. The numerator tan³x+tan x equals t(1+t²), so \(\int\frac{\tan^3x+\tan x}{\tan x+4}\,dx=\int\frac{t}{t+4}\,dt.\) Polynomial division gives t/(t+4)=1−4/(t+4). Integrate and return to x. Work on intervals where tan x exists and tan x≠−4.
Final result
\[\tan x-4\log|\tan x+4|+C\]Exercise 5
Solve the following exercise:
\[\int_{4}^{16} \frac{1}{(x-\sqrt{x})^2} \, dx.\]Solution.
Set t=√x, dx=2t dt. The bounds become 2 and 4, and (x−√x)²=t²(t−1)². Hence \(I=\int_2^4\frac2{t(t-1)^2}\,dt.\) Determine the partial fractions: \(\frac2{t(t-1)^2}=\frac2t-\frac2{t-1}+\frac2{(t-1)^2}.\) A primitive is 2log t−2log(t−1)−2/(t−1). Its values at 4 and 2 give \(I=2\log\frac43-\frac23-(2\log2-2).\)
Final result
\[\frac43+2\log\frac23\]Exercise 6
Solve the following exercise:
\[\int \frac{2 \tan x+1}{\sin^{2} x +3 \cos^{2} x} \, dx\]Solution.
Set t=tan x, so dx=dt/(1+t²), sin²x=t²/(1+t²) and cos²x=1/(1+t²). The resulting rational integral is \(\int\frac{2t+1}{t^2+3}\,dt=\int\frac{2t}{t^2+3}\,dt+\int\frac{dt}{t^2+3}.\) The first term is a logarithmic derivative; rescale t by √3 in the second to obtain an arctangent. This substitution works on intervals where tan x is defined.
Final result
\[\log(\tan^2x+3)+\frac1{\sqrt3}\arctan\frac{\tan x}{\sqrt3}+C\]Exercise 7
Evaluate the integral:
\[\int \frac{1}{x^{3}+1} \, dx.\]Solution.
This original statement appears in the unsolved-exercise list. The following worked solution is supplied for this edition.
Factor x³+1=(x+1)(x²−x+1) and determine coefficients: \(\frac1{x^3+1}=\frac1{3(x+1)}+\frac{-x+2}{3(x^2-x+1)}.\) In the quadratic numerator use −x+2=−(2x−1)/2+3/2. The derivative part integrates to −log(x²−x+1)/6. Complete the square x²−x+1=(x−1/2)²+3/4 to integrate the remaining part as an arctangent. The pole x=−1 separates the primitive intervals.
Final result
\[\frac13\log|x+1|-\frac16\log(x^2-x+1)+\frac1{\sqrt3}\arctan\frac{2x-1}{\sqrt3}+C\]Exercise 8
Evaluate the integral:
\[\int \frac{x-1}{x^{2}-4x+5} \, dx.\]Solution.
This original statement appears in the unsolved-exercise list. The following worked solution is supplied for this edition.
Write the denominator as (x−2)²+1, and split x−1=(x−2)+1. Then \(\int\frac{x-1}{(x-2)^2+1}\,dx=\int\frac{x-2}{(x-2)^2+1}\,dx+\int\frac{dx}{(x-2)^2+1}.\) The first term is half the logarithmic derivative of the denominator; the second is the derivative of arctan(x−2). The denominator is positive for every real x.
Final result
\[\frac12\log(x^2-4x+5)+\arctan(x-2)+C\]