How do you integrate int (x^3 - 2) / (x^4 - 1)∫x3−2x4−1 using partial fractions?
1 Answer
Explanation:
First, let's factor
x^4-1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)x4−1=(x2−1)(x2+1)=(x−1)(x+1)(x2+1)
Using partial fraction separation, we can say that:
(x^3-2)/((x-1)(x+1)(x^2+1)) = A/(x-1) + B/(x+1)+(Cx+D)/(x^2+1)x3−2(x−1)(x+1)(x2+1)=Ax−1+Bx+1+Cx+Dx2+1
x^3-2 = A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x+1)(x-1)x3−2=A(x+1)(x2+1)+B(x−1)(x2+1)+(Cx+D)(x+1)(x−1)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Both the
(-1)^3-2 = A(0)(2)+B(-2)(2)+(-C+D)(0)(-2)(−1)3−2=A(0)(2)+B(−2)(2)+(−C+D)(0)(−2)
-3 = -4B−3=−4B
3/4 = B34=B
Now we can substitute this into the original equation.
x^3-2 = A(x+1)(x^2+1)+3/4(x-1)(x^2+1)+(Cx+D)(x+1)(x-1)x3−2=A(x+1)(x2+1)+34(x−1)(x2+1)+(Cx+D)(x+1)(x−1)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute
1-2 = A(2)(2) + 3/4(0)(2) + (C+D)(2)(0)1−2=A(2)(2)+34(0)(2)+(C+D)(2)(0)
-1 = 4A−1=4A
-1/4 = A−14=A
Now substitute this into the original equation.
x^3-2 = -1/4(x+1)(x^2+1)+3/4(x-1)(x^2+1)+(Cx+D)(x+1)(x-1)x3−2=−14(x+1)(x2+1)+34(x−1)(x2+1)+(Cx+D)(x+1)(x−1)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Plug in
-2 = -1/4(1)(1)+3/4(-1)(1)+(0C+D)(1)(-1)−2=−14(1)(1)+34(−1)(1)+(0C+D)(1)(−1)
-2 = -1/4 - 3/4 - D−2=−14−34−D
-1 = -D−1=−D
1 = D1=D
Now substitute this into the original equation:
x^3-2 = -1/4(x+1)(x^2+1)+3/4(x-1)(x^2+1)+(Cx+1)(x+1)(x-1)x3−2=−14(x+1)(x2+1)+34(x−1)(x2+1)+(Cx+1)(x+1)(x−1)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now we can pretty much plug in any number to solve for
2^3-2 = -1/4(3)(5)+3/4(1)(5)+(2C+1)(3)(1)23−2=−14(3)(5)+34(1)(5)+(2C+1)(3)(1)
6 = -15/4+15/4 + (6C+3)6=−154+154+(6C+3)
6 = 6C+36=6C+3
3=6C3=6C
1/2 = C12=C
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So we have:
(x^3-2)/((x-1)(x+1)(x^2+1)) = (-1/4)/(x-1) + (3/4)/(x+1)+(1/2x+1)/(x^2+1)x3−2(x−1)(x+1)(x2+1)=−14x−1+34x+1+12x+1x2+1
All that is left to do is take the integral of this function.
int[(-1/4)/(x-1) + (3/4)/(x+1)+(1/2x+1)/(x^2+1)]dx∫[−14x−1+34x+1+12x+1x2+1]dx
= int(-1/4)/(x-1)dx + int(3/4)/(x+1)dx + int(1/2x+1)/(x^2+1)dx=∫−14x−1dx+∫34x+1dx+∫12x+1x2+1dx
We can integrate the first two pretty easily, and then we can split the numerator of the last integral into two.
= -1/4ln|x-1|+3/4ln|x+1|+int(1/2x)/(x^2+1)dx + int1/(x^2+1)dx=−14ln|x−1|+34ln|x+1|+∫12xx2+1dx+∫1x2+1dx
Use the substitution
= -1/4ln|x-1|+3/4ln|x+1|+int(1/4)/udu + int1/(x^2+1)dx=−14ln|x−1|+34ln|x+1|+∫14udu+∫1x2+1dx
= -1/4ln|x-1|+3/4ln|x+1|+1/4ln(u) + tan^-1(x)=−14ln|x−1|+34ln|x+1|+14ln(u)+tan−1(x)
And finally simplify everything (don't forget
= -1/4ln|x-1|+1/4ln|(x+1)^3| + 1/4ln|x^2+1| + tan^-1(x)=−14ln|x−1|+14ln∣∣(x+1)3∣∣+14ln∣∣x2+1∣∣+tan−1(x)
= 1/4ln|((x^2+1)(x+1)^3)/(x-1)|+tan^-1(x) + C=14ln∣∣ ∣∣(x2+1)(x+1)3x−1∣∣ ∣∣+tan−1(x)+C
Final Answer
