Question

How do you integrate \[\dfrac{{{x^4} + 1}}{{{x^3} + 2x}}\] using partial fractions?

Answer 1

Hint: We need to integrate the given equation by using the partial fractions. In order to use the partial fractions the power value of the numerator must be less than the power value of the denominator. Otherwise, we have to use a long division method to solve the problem.

Complete step-by-step answer:
Given,
Integrate \[\dfrac{{{x^4} + 1}}{{{x^3} + 2x}}\]
By using long division method, we get
 \[\dfrac{{{x^4} + 1}}{{{x^3} + 2x}} = x - \dfrac{{2{x^2} - 1}}{{{x^3} + 2x}}\]
 Take factoring out \[x\] from the denominator, we get
 \[\dfrac{{2{x^2} - 1}}{{{x^3} + 2x}} = x - \dfrac{{2{x^2} - 1}}{{x({x^2} + 2)}}\]
By using the partial decomposition of \[\dfrac{{2{x^2} - 1}}{{x({x^2} + 2)}}\] and break into multiple factors,
 \[\dfrac{{2{x^2} - 1}}{{x({x^2} + 2)}} = \dfrac{A}{x} + \dfrac{{Bx}}{{{x^2} + 2}}\] …………… \[(1)\]
By multiplying common denominator \[x({x^2} + 2)\] on both sides, we get
 \[2{x^2} - 1 = A({x^2} + 2) + B{x^2}\] ………. \[(2)\]
To find the value of \[A\] and \[B\] by the factor value of \[x = 0,\] then put \[x = 1\] ,
If \[x = 0,\] substitute the equation \[(2)\] gives,
  \[
  2(0) - 1 = A(0 + 2) + B(0) \\
   - 1 = 2A \\
  A = \dfrac{{ - 1}}{2} \\
\] \[\]
By applying the value \[A\] in equation \[(2)\]
 \[2{x^2} - 1 = - \dfrac{1}{2}({x^2} + 2) + B{x^2}\] …………. \[(3)\] \[\]
If \[x = 1\] , substitute in equation \[(3)\] gives,
 \[
  1 = - \dfrac{3}{2} + B \;
 \]
To simplify, then we get
 \[B = \dfrac{{2 + 3}}{2}\]
Then, the value of \[B = \dfrac{5}{2}\]
By substitute all the values into the equation \[(1)\] , Partial decomposition
 \[
  \dfrac{{2{x^2} - 1}}{{x({x^2} + 2)}} = x - \dfrac{{\left[ {\dfrac{{ - 1}}{2}} \right] }}{x} + \dfrac{{\dfrac{5}{2}x}}{{{x^2} + 2}} \\
  \dfrac{{{x^4} - 1}}{{({x^3} + 2x)}} = x + \dfrac{{\dfrac{1}{2}}}{x} + \dfrac{{\dfrac{5}{2}x}}{{{x^2} + 2}} \;
 \]
To integrating the above equation, we get
 \[
  \int {\left( {x + \dfrac{{\dfrac{1}{2}}}{x} + \dfrac{{\dfrac{5}{2}x}}{{{x^2} + 2}}} \right)dx} \\
  \int {xdx + \int {\dfrac{{\dfrac{1}{2}}}{x}dx + \dfrac{5}{2}\int {\dfrac{x}{{{x^2} + 2}}dx} } } \;
 \]
To simplify the integration,
 \[\int {xdx + \dfrac{1}{2}\int {\dfrac{1}{x}dx + \dfrac{5}{2}\int {\left( {\dfrac{x}{{{x^2} + 2}}} \right)dx} } } \]
By apply the integral formula, we have
 \[\dfrac{{{x^2}}}{2} + \dfrac{1}{2}In(x) + C + \dfrac{5}{2}\int {\left( {\dfrac{x}{{{x^2} + 2}}} \right)dx} \] …………. \[(4)\]
By substitute \[u\] on the last equation of above, we get
Let us consider,
 \[u = {x^2} + 2\]
By differentiate \[u\] with respect to \[x\] , we get
 \[
  du = 2xdx \\
  xdx = \dfrac{1}{2}du \;
 \]
Now, \[\int {\left( {\dfrac{x}{{{x^2} + 2}}} \right)dx} = \int {\left( {\dfrac{1}{u}} \right)\dfrac{1}{2}du = \dfrac{1}{2}In(u) + C} \]
So, apply back the value of \[u = {x^2} + 2\] ,
 \[\int {\left( {\dfrac{x}{{{x^2} + 2}}} \right)dx} = \dfrac{1}{2}In({x^2} + 2) + C\] ……….. \[(5)\]
By substitute the equation \[(5)\] in \[(4)\] , we get
 \[\dfrac{{{x^2}}}{2} + \dfrac{1}{2}In(x) + C + \dfrac{5}{2}\left( {\dfrac{1}{2}In({x^2} + 2) + C} \right)\]
Hence, The final answer is \[\dfrac{{{x^2}}}{2} + \dfrac{1}{2}In(x) + C + \dfrac{5}{2}In\left( {{x^2} + 2} \right) + C\] .
So, the correct answer is “\[\dfrac{{{x^2}}}{2} + \dfrac{1}{2}In(x) + \dfrac{5}{2}In\left( {{x^2} + 2} \right) + C\] ”.

Note: We need to integrate the given equation by using the partial fractions. In order to use the partial fractions the power value of the numerator must be less than the power value of the denominator. Otherwise, we have to use a long division method to solve the problem and use some integration formula.