How can I integrate #int 1/(x^4+x) dx# using partial fractions?
My problem is that I don't know what will be in the numerator over the #x^3+1#
#A# or #Ax+B# ? I've learnt that only these two options exist. But I also saw #Ax^2+B# so I'm kinda confused.
My problem is that I don't know what will be in the numerator over the
2 Answers
Explanation:
Now I decomposed integrand into basic fractions,
=
=
=
=
=
The answer is
Explanation:
Reminder
The denominator is
Therefore, the decomposition into partial fractions is
The denominators are the same, compare the numerators
Compare the LHS and the RHS
Let
Let
Coefficients of
Coefficients of
Therefore,
So,