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Q11:
Integrate the rational functions in Exercises 1 to 21. 11. $\frac{5x}{(x+1)(x^2-4)}$

Solution :

Step 1: Decomposition of the Rational Function

To integrate $\int \frac{5x}{(x+1)(x^2-4)} dx$, we first factorize the quadratic term in the denominator. Recall the algebraic identity $a^2 - b^2 = (a-b)(a+b)$.

$x^2 - 4 = (x-2)(x+2)$

Therefore, the integrand becomes:

$\frac{5x}{(x+1)(x-2)(x+2)}$

[By Partial Fraction Decomposition theorem, since all factors are linear and distinct, we assume]:

$\frac{5x}{(x+1)(x-2)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+2}$

Step 2: Determining the Constants A, B, and C

Multiply both sides by $(x+1)(x-2)(x+2)$ to obtain the identity:

$5x = A(x-2)(x+2) + B(x+1)(x+2) + C(x+1)(x-2)$

$5x = A(x^2 - 4) + B(x^2 + 3x + 2) + C(x^2 - x - 2)$

  • To find $A$, set $x = -1$: $5(-1) = A(-1-2)(-1+2) \implies -5 = A(-3)(1) \implies A = \frac{5}{3}$
  • To find $B$, set $x = 2$: $5(2) = B(2+1)(2+2) \implies 10 = B(3)(4) \implies 10 = 12B \implies B = \frac{10}{12} = \frac{5}{6}$
  • To find $C$, set $x = -2$: $5(-2) = C(-2+1)(-2-2) \implies -10 = C(-1)(-4) \implies -10 = 4C \implies C = -\frac{10}{4} = -\frac{5}{2}$

Step 3: Integration of the Partial Fractions

The integral can now be expressed as the sum of three elementary integrals:

$\int \frac{5x}{(x+1)(x^2-4)} dx = \frac{5}{3} \int \frac{1}{x+1} dx + \frac{5}{6} \int \frac{1}{x-2} dx - \frac{5}{2} \int \frac{1}{x+2} dx$

[Applying the standard integral rule $\int \frac{1}{x} dx = \ln|x| + C$]:

$= \frac{5}{3} \ln|x+1| + \frac{5}{6} \ln|x-2| - \frac{5}{2} \ln|x+2| + C$

Visual Representation of Partial Fraction Components

Integrand: 5x / [(x+1)(x-2)(x+2)] A/(x+1) = (5/3)/(x+1) B/(x-2) = (5/6)/(x-2) C/(x+2) = -(5/2)/(x+2)

Step 4: Simplification

To provide a more compact form, we observe the coefficients $\frac{5}{3}, \frac{5}{6}, -\frac{5}{2}$. Finding a common denominator ($6$):

$= \frac{10}{6} \ln|x+1| + \frac{5}{6} \ln|x-2| - \frac{15}{6} \ln|x+2| + C$

$= \frac{5}{6} [2\ln|x+1| + \ln|x-2| - 3\ln|x+2|] + C$

[Using logarithmic power rule $n\ln(a) = \ln(a^n)$ and quotient rule $\ln(a) - \ln(b) = \ln(a/b)$]:

$= \frac{5}{6} \ln \left| \frac{(x+1)^2(x-2)}{(x+2)^3} \right| + C$

Final Solution: \frac{5}{3} \ln|x+1| + \frac{5}{6} \ln|x-2| - \frac{5}{2} \ln|x+2| + C


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