Solveeit Logo
Login

Question

Question: How do you integrate \(\int {\dfrac{{x\left( {x + 2} \right)}}{{{x^3} + 3{x^2} - 4}}dx} \)?...

How do you integrate ∫x(x+2)x3+3x2−4dx\int {\dfrac{{x\left( {x + 2} \right)}}{{{x^3} + 3{x^2} - 4}}dx} ?

Explanation

Solution

Here in this question, we have to find the integrated value of a given function by using partial fractions method. I would try Integration by Partial Fractions when an integrand is a rational function with its denominator can be factored out into smaller factors and by further simplifying using the standard integration formula. To get the required solution.

Complete step by step answer:
If the integrand (the expression after the integral sign) is in the form of an algebraic fraction and the integral cannot be evaluated by simple methods, the fraction needs to be expressed in partial fractions before integration takes place.The steps needed to decompose an algebraic fraction into its partial fractions results from a consideration of the reverse process − addition (or subtraction).Consider the given integral function:
∫x(x+2)x3+3x2−4dx\int {\dfrac{{x\left( {x + 2} \right)}}{{{x^3} + 3{x^2} - 4}}dx} ---------(1)
Let us factorize the denominator.
It is easy to see that x3+3x2−4{x^3} + 3{x^2} - 4 vanishes when x=1x = 1, so that (x−1)\left( {x - 1} \right) is a factor. Then,
⇒  x3+3x2−4\Rightarrow \,\,{x^3} + 3{x^2} - 4
⇒  x3−x2+4x2−4x+4x−4\Rightarrow \,\,{x^3} - {x^2} + 4{x^2} - 4x + 4x - 4

Take out greatest common divisor GCD, then
⇒  x3−x2+4x2−4x+4x−4\Rightarrow \,\,{x^3} - {x^2} + 4{x^2} - 4x + 4x - 4
⇒  x2(x−1)+4x(x−1)+4(x−1)\Rightarrow \,\,{x^2}\left( {x - 1} \right) + 4x\left( {x - 1} \right) + 4\left( {x - 1} \right)
Take (x−1)\left( {x - 1} \right) as common
⇒  (x−1)(x2+4x+4)\Rightarrow \,\,\left( {x - 1} \right)\left( {{x^2} + 4x + 4} \right)
By the algebraic identity (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}.
⇒  (x−1)(x+2)2\Rightarrow \,\,\left( {x - 1} \right){\left( {x + 2} \right)^2} --- (2)
Substitute equation (2) in denominator of equation (1)
⇒   ∫x(x+2)(x−1)(x+2)2dx\Rightarrow \,\,\,\int {\dfrac{{x\left( {x + 2} \right)}}{{\left( {x - 1} \right){{\left( {x + 2} \right)}^2}}}dx}
On simplification, we have
⇒   ∫x(x−1)(x+2)dx\Rightarrow \,\,\,\int {\dfrac{x}{{\left( {x - 1} \right)\left( {x + 2} \right)}}dx} ------(3)

Now, separate the fraction that we wish to decompose into multiple fractions. The factor of xx in the denominator has a power higher than 1, then the coefficients in the numerator should reflect this higher power.To use the Method of Partial Fractions, we let, for,   A,B∈R\;A,B \in R, then equation (3) can be written as
⇒   x(x−1)(x+2)=A(x−1)+B(x+2)\Rightarrow \,\,\,\dfrac{x}{{\left( {x - 1} \right)\left( {x + 2} \right)}} = \dfrac{A}{{\left( {x - 1} \right)}} + \dfrac{B}{{\left( {x + 2} \right)}}--------(4)
Take (x−1)(x+2)\left( {x - 1} \right)\left( {x + 2} \right) as LCM in RHS, then
⇒   x=A(x+2)+B(x−1)\Rightarrow \,\,\,x = A\left( {x + 2} \right) + B\left( {x - 1} \right)----------(5)
AA can easily be obtained by using Heavyside's Cover-up Method, The Heaviside cover-up method, is one possible approach in determining the coefficients when performing the partial-fraction expansion of a rational function.
To find AA, Put x=1x = 1in equation (5)
⇒   1=A(1+2)+B(1−1)\Rightarrow \,\,\,1 = A\left( {1 + 2} \right) + B\left( {1 - 1} \right)
⇒   1=A(1+2)+B(0)\Rightarrow \,\,\,1 = A\left( {1 + 2} \right) + B\left( 0 \right)
⇒   1=A(3)+0\Rightarrow \,\,\,1 = A\left( 3 \right) + 0

Divide both side by 3, then
∴   A=13\therefore \,\,\,A = \dfrac{1}{3}
To find BB, Put x=−2x = - 2in equation (5)
⇒   −2=A(−2+2)+B(−2−1)\Rightarrow \,\,\, - 2 = A\left( { - 2 + 2} \right) + B\left( { - 2 - 1} \right)
⇒   −2=A(0)+B(−3)\Rightarrow \,\,\, - 2 = A\left( 0 \right) + B\left( { - 3} \right)
⇒   −2=0−3B\Rightarrow \,\,\, - 2 = 0 - 3B
⇒   −2=−3B\Rightarrow \,\,\, - 2 = - 3B
Divide both side by -3, then
∴   B=23\therefore \,\,\,B = \dfrac{2}{3}
Substitute the value of AA and BBin equation (4):
⇒   x(x−1)(x+2)=(13)(x−1)+(23)(x+2)\Rightarrow \,\,\,\dfrac{x}{{\left( {x - 1} \right)\left( {x + 2} \right)}} = \dfrac{{\left( {\dfrac{1}{3}} \right)}}{{\left( {x - 1} \right)}} + \dfrac{{\left( {\dfrac{2}{3}} \right)}}{{\left( {x + 2} \right)}}

It can be written as
⇒   x(x−1)(x+2)=13(x−1)+23(x+2)\Rightarrow \,\,\,\dfrac{x}{{\left( {x - 1} \right)\left( {x + 2} \right)}} = \dfrac{1}{{3\left( {x - 1} \right)}} + \dfrac{2}{{3\left( {x + 2} \right)}}
Integrate each fraction with respect to xx
⇒   ∫x(x−1)(x+2) dx=∫13(x−1) dx+∫23(x+2) dx\Rightarrow \,\,\,\int {\dfrac{x}{{\left( {x - 1} \right)\left( {x + 2} \right)}}} \,dx = \int {\dfrac{1}{{3\left( {x - 1} \right)}}} \,dx + \int {\dfrac{2}{{3\left( {x + 2} \right)}}} \,dx
⇒   ∫x(x−1)(x2+4) dx=13ln⁡∣x−1∣+23ln⁡∣x+2∣+K\Rightarrow \,\,\,\int {\dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}}} \,dx = \dfrac{1}{3}\ln \left| {x - 1} \right| + \dfrac{2}{3}\ln \left| {x + 2} \right| + K
∴   ∫x(x+2)x3+3x2−4 dx=13ln⁡∣x−1∣+23ln⁡∣x+2∣+K\therefore \,\,\,\int {\dfrac{{x\left( {x + 2} \right)}}{{{x^3} + 3{x^2} - 4}}\,dx} = \dfrac{1}{3}\ln \left| {x - 1} \right| + \dfrac{2}{3}\ln \left| {x + 2} \right| + K
Where KK is an integrating constant.

Note: When the function is in the form of a fraction where the denominator of the function is in the form of a polynomial, then we use the partial fractions and then we integrate the function. The integration is a reciprocal of the differentiation and by using the standard formulas of integration we determine the values.