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Question: Solve the following \(\int {\dfrac{{{x^3} - x}}{{1 + {x^6}}}dx{\text{ }}} \)...

Solve the following
∫x3−x1+x6dx \int {\dfrac{{{x^3} - x}}{{1 + {x^6}}}dx{\text{ }}}

Explanation

Solution

Hint: In this question, we have to evaluate the given equation by using integral and its properties and some rules but to solve this question we’ll remarkably work on the typical methods for solving this integral equation. Thus, we will be able to simplify this equation and hence we can get a desirable result.

Complete step-by-step answer:
In order to solve this, first we will factorise the denominator
So, we have to make factors of the denominator i.e.1+x61 + {x^6}
Now, substitute x2=y{x^2} = y
We will get 1+y31 + {y^3}, notice that y=−1y = - 1 is a solution of 1+y3=01 + {y^3} = 0
So, (y+1)\left( {y + 1} \right) is a factor and also
By using the algebraic formula i.e. a3+b3=(a+b)(a2+b2−ab){a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)
We get 1+y3=(1+y)(1+y2−y)1 + {y^3} = \left( {1 + y} \right)\left( {1 + {y^2} - y} \right)
Substituting y=x2y = {x^2} back in, we get
1+x6=(1+x2)(1+x4−x2)1 + {x^6} = \left( {1 + {x^2}} \right)\left( {1 + {x^4} - {x^2}} \right)
Therefore, the equation changes to ∫x3−x(x2+1)(x4−x2+1)dx \int {\dfrac{{{x^3} - x}}{{\left( {{x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right)}}dx{\text{ }}}
Now, we have to perform the partial decomposition
x3−x(x2+1)(x4−x2+1)=Ax+Bx2+1+Cx3+Dx2+Ex+Fx4−x2+1\dfrac{{{x^3} - x}}{{\left( {{x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right)}} = \dfrac{{Ax + B}}{{{x^2} + 1}} + \dfrac{{C{x^3} + D{x^2} + Ex + F}}{{{x^4} - {x^2} + 1}} … (1)
Multiply the whole equation with (x2+1)(x4−x2+1)\left( {{x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right)
x3−x=(Ax+B)(x4−x2+1)+(Cx3+Dx2+Ex+F)(x2+1)  =Ax5−Ax3+Ax+Bx4−Bx2+B+Cx5+Cx3+Dx4+Dx2+Ex3+Ex+Fx2+F   {x^3} - x = \left( {Ax + B} \right)\left( {{x^4} - {x^2} + 1} \right) + \left( {C{x^3} + D{x^2} + Ex + F} \right)\left( {{x^2} + 1} \right) \\\ {\text{ }} = A{x^5} - A{x^3} + Ax + B{x^4} - B{x^2} + B + C{x^5} + C{x^3} + D{x^4} + D{x^2} + E{x^3} + Ex + F{x^2} + F \\\ \\\
Group them according to power of xx,
x3−x=x5(A+C)+x4(B+D)+x3(−A+C+E)+x2(−B+D+F)+x(A+E)+B+F   {x^3} - x = {x^5}\left( {A + C} \right) + {x^4}\left( {B + D} \right) + {x^3}\left( { - A + C + E} \right) + {x^2}\left( { - B + D + F} \right) + x\left( {A + E} \right) + B + F \\\ \\\
Equate the coefficients of similar terms,
For x5: A+C=0 ∴C=−A For x4: B+D=0 ∴D=−B For x3: \-A+C+E=1 ...(2) For x2: \-B+D+F=0 ...(3) For x: A+E=−1 ∴E=−1−A Constants or for x0: B+F=0 ∴F=−B  {\text{For }}{x^5}: \\\ A + C = 0 \\\ \therefore C = - A \\\ {\text{For }}{x^4}: \\\ B + D = 0 \\\ \therefore D = - B \\\ {\text{For }}{x^3}: \\\ \- A + C + E = 1{\text{ }}...\left( 2 \right) \\\ {\text{For }}{x^2}: \\\ \- B + D + F = 0{\text{ }}...\left( 3 \right) \\\ {\text{For }}x: \\\ A + E = - 1 \\\ \therefore E = - 1 - A \\\ {\text{Constants or for }}{x^0}: \\\ B + F = 0 \\\ \therefore F = - B \\\
Now, put the values of A and E in eq.(2) and D and F in eq.(3)A{\text{ and }}E{\text{ in eq}}{\text{.}}\left( 2 \right){\text{ and }}D{\text{ and }}F{\text{ in eq}}{\text{.}}\left( 3 \right)
⇒−A−A−1−A=1 and −B−B−B=0 ⇒−3A−1=1 and −3B=0 ⇒A=−23 and B=0  \Rightarrow - A - A - 1 - A = 1{\text{ and }} - B - B - B = 0 \\\ \Rightarrow - 3A - 1 = 1{\text{ and }} - 3B = 0 \\\ \Rightarrow A = - \dfrac{2}{3}{\text{ and }}B = 0 \\\
Now, put the values of A and BA{\text{ and }}B back in order to find the remaining the values
Therefore, C=−A=23 D=−B=0 E=−1−A=−1+23=−13 F=−B=0  C = - A = \dfrac{2}{3} \\\ D = - B = 0 \\\ E = - 1 - A = - 1 + \dfrac{2}{3} = \dfrac{{ - 1}}{3} \\\ F = - B = 0 \\\
Substitute all the variables back in eq. (1)
x3−x(x2+1)(x4−x2+1)=(−23)x+0x2+1+(23)x3+(0)x2+(−13)x+0x4−x2+1  =−2x3(x2+1)+2x3−x3(x4−x2+1)  \dfrac{{{x^3} - x}}{{\left( {{x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right)}} = \dfrac{{\left( {\dfrac{{ - 2}}{3}} \right)x + 0}}{{{x^2} + 1}} + \dfrac{{\left( {\dfrac{2}{3}} \right){x^3} + \left( 0 \right){x^2} + \left( {\dfrac{{ - 1}}{3}} \right)x + 0}}{{{x^4} - {x^2} + 1}} \\\ {\text{ }} = \dfrac{{ - 2x}}{{3\left( {{x^2} + 1} \right)}} + \dfrac{{2{x^3} - x}}{{3\left( {{x^4} - {x^2} + 1} \right)}} \\\
Again, integrating both sides

∫x3−x(x2+1)(x4−x2+1)dx=∫−2x3(x2+1)dx+∫2x3−x3(x4−x2+1)dx  =−23∫xx2+1dx+13∫2x3−xx4−x2+1dx   \int {\dfrac{{{x^3} - x}}{{\left( {{x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right)}}dx = \int {\dfrac{{ - 2x}}{{3\left( {{x^2} + 1} \right)}}dx + \int {\dfrac{{2{x^3} - x}}{{3\left( {{x^4} - {x^2} + 1} \right)}}dx} } } \\\ {\text{ }} = \dfrac{{ - 2}}{3}\int {\dfrac{x}{{{x^2} + 1}}dx} + \dfrac{1}{3}\int {\dfrac{{2{x^3} - x}}{{{x^4} - {x^2} + 1}}dx} {\text{ }} \\\
Now, solving ∫2x3−xx4−x2+1dx\int {\dfrac{{2{x^3} - x}}{{{x^4} - {x^2} + 1}}dx}
substitute u=x4−x2+1→dudx=4x3−2x→dx=14x3−2xdu   {\text{substitute }}u = {x^4} - {x^2} + 1 \to \dfrac{{du}}{{dx}} = 4{x^3} - 2x \to dx = \dfrac{1}{{4{x^3} - 2x}}du \\\ \\\
⇒∫2x3−xx4−x2+1dx=∫2x3−xu×14x3−2xdu=∫12udu=12∫1udu\Rightarrow \int {\dfrac{{2{x^3} - x}}{{{x^4} - {x^2} + 1}}dx} = \int {\dfrac{{2{x^3} - x}}{u} \times \dfrac{1}{{4{x^3} - 2x}}du} = \int {\dfrac{1}{{2u}}du = \dfrac{1}{2}\int {\dfrac{1}{u}du} }
As you know this is a standard integral
So, 12∫1udu=ln⁡(u)2\dfrac{1}{2}\int {\dfrac{1}{u}du} = \dfrac{{\ln \left( u \right)}}{2}
Put the value back in uu we get,
ln⁡(x4−x2+1)2\dfrac{{\ln \left( {{x^4} - {x^2} + 1} \right)}}{2} … (4)
Now, solving ∫xx2+1dx\int {\dfrac{x}{{{x^2} + 1}}dx}
substitute u=x2+1→dudx=2x→dx=12xdu   {\text{substitute }}u = {x^2} + 1 \to \dfrac{{du}}{{dx}} = 2x \to dx = \dfrac{1}{{2x}}du \\\ \\\
⇒∫xx2+1dx=∫xu×12xdu=12∫1udu\Rightarrow \int {\dfrac{x}{{{x^2} + 1}}dx} = \int {\dfrac{x}{u} \times \dfrac{1}{{2x}}du = \dfrac{1}{2}\int {\dfrac{1}{u}du} }
Similarly, as above
12∫1udu=ln⁡(u)2\dfrac{1}{2}\int {\dfrac{1}{u}du} = \dfrac{{\ln \left( u \right)}}{2}
Again put the value back in uu we get,
ln⁡(x2+1)2\dfrac{{\ln \left( {{x^2} + 1} \right)}}{2} … (4)
Substitute these answers i.e. eq. (5) and eq. (6) back in the original eq.
−23∫xx2+1dx+13∫2x3−xx4−x2+1dx=13×ln⁡(x4−x2+1)2−23×ln⁡(x2+1)2=ln⁡(x4−x2+1)6−ln⁡(x2+1)3\dfrac{{ - 2}}{3}\int {\dfrac{x}{{{x^2} + 1}}dx} + \dfrac{1}{3}\int {\dfrac{{2{x^3} - x}}{{{x^4} - {x^2} + 1}}dx} = \dfrac{1}{3} \times \dfrac{{\ln \left( {{x^4} - {x^2} + 1} \right)}}{2} - \dfrac{2}{3} \times \dfrac{{\ln \left( {{x^2} + 1} \right)}}{2} = \dfrac{{\ln \left( {{x^4} - {x^2} + 1} \right)}}{6} - \dfrac{{\ln \left( {{x^2} + 1} \right)}}{3}
Hence, the problem is solved
∫x3−x1+x6dx =ln⁡(x4−x2+1)6−ln⁡(x2+1)3+C=ln⁡(x4−x2+1)−2ln⁡(x2+1)6+C\int {\dfrac{{{x^3} - x}}{{1 + {x^6}}}dx{\text{ }}} = \dfrac{{\ln \left( {{x^4} - {x^2} + 1} \right)}}{6} - \dfrac{{\ln \left( {{x^2} + 1} \right)}}{3} + C = \dfrac{{\ln \left( {{x^4} - {x^2} + 1} \right) - 2\ln \left( {{x^2} + 1} \right)}}{6} + C where, CC is integral constant.

Note: It is to be noted that, since this question includes innumerable integral properties and rules but to solve this question easily and quickly we must know the integral methods too and focus on them eminently as they play vital roles in these types of questions and these methods vary accordingly to the particular question. As in this question, first we use factorisation for denominators, then we use partial decomposition and find the values of all the variables used. After substituting the values of variables we use a substitution method in order to simplify the equation and after evaluating those two expressions one by one. We must put it back in the original equation to solve the given question and get our required result.