Before we rewrite x^4/(x^4-1)x4x4−1 into partial fractions, first make sure that the degree of the denominator is strictly greater than the degree of the numerator.
So, we can use x^4/(x^4-1)=(x^4-1+1)/(x^4-1)=1+1/(x^4-1)x4x4−1=x4−1+1x4−1=1+1x4−1
Now, we rewrite 1/(x^4-1)1x4−1 using partial fractions. First, factorize x^4-1=(x^2+1)(x+1)(x-1)x4−1=(x2+1)(x+1)(x−1).
Thus, we know that 1/(x^4-1)1x4−1 can be rewritten in the form (Ax+B)/(x^2+1)+C/(x+1)+D/(x-1)Ax+Bx2+1+Cx+1+Dx−1.
Multiply both sides by x^4-1x4−1 to get 1=(Ax+B)(x+1)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1)1=(Ax+B)(x+1)(x−1)+C(x2+1)(x−1)+D(x2+1)(x+1).
Set x=1x=1 to get D=1/4D=14 and x=-1x=−1 to get C=-1/4C=−14
Multiply this out to get (A+C+D)x^3+(B-C+D)x^2-(A-C-D)x-B-C+D=1(A+C+D)x3+(B−C+D)x2−(A−C−D)x−B−C+D=1.
Substitute D=1/4D=14 and C=-1/4C=−14 to finally get Ax^3+(B+1/2)x^2-Ax-B+1/2=1Ax3+(B+12)x2−Ax−B+12=1.
Equate coefficients: {(A=0),(B+1/2=0),(A=0),(-B+1/2=1):}.
From the above, we get {(A=0),(B=-1/2),(C=-1/4),(D=1/4):}.
Thus, we know that 1/(x^4-1)=-1/(2x^2+2)-1/(4x+4)+1/(4x-4).
Returning to the original problem, we can now integrate int\ 1-1/(2x^2+2)-1/(4x+4)+1/(4x-4)\ dx.
The first, third, and fourth can be easily solved using either the constant rule or u-substitution: int\ dx=x, -int\ dx/(4x+4)=-(ln|x+1|)/4, and int\ dx/(4x-4)=(ln|x-1|)/4.
The second one, -int\ dx/(2x^2+2)=-1/2int\ dx/(x^2+1), requires knowing the identity int\ dx/(x^2+1)=arctan(x).
Thus, we can combine everything to get the final answer: x+1/4ln|(x-1)/(x+1)|-arctan(x)/2+C.
