### Video Transcript

Use partial fractions to evaluate
the indefinite integral of one over π‘ cubed plus π‘ squared minus two π‘ with
respect to π‘.

When rewriting the integrand using
partial fractions, weβll be looking to reverse the process we would take when adding
algebraic fractions. Before we can do this though, we
need to decide what the denominator of these fractions might be. So letβs see if we can factor the
denominator.

Itβs clearly a cubic, but thereβs a
common factor of π‘. So weβll factor the denominator by
π‘ to get π‘ times π‘ squared plus π‘ minus two. We can then factor this quadratic
expression the usual way. π‘ squared plus π‘ minus two is
equal to π‘ minus one times π‘ plus two. We can now split this into partial
fractions. These are π΄ over π‘ plus some
other constant π΅ over π‘ minus one plus πΆ over π‘ plus two.

Next, weβre going to add these
three fractions. Weβll need to multiply the
numerator and denominator of each of them by the denominator of the other two. So the numerator becomes π΄ times
π‘ minus one times π‘ plus two plus π΅ times π‘ times π‘ plus two plus πΆ times π‘
times π‘ minus one. And now we notice that the
denominator of the fractions on both sides of this equation are equal. And for the fractions themselves to
be equal, this means in turn that their numerators must be equal. So we can say that one is equal to
π΄ times π‘ minus one times π‘ plus two plus π΅π‘ times π‘ plus two plus πΆπ‘ times
π‘ minus one. Weβre going to use the method of
substitution to help us find the values for π΄, π΅, and πΆ.

Notice that if we let π‘ be equal
to zero, we can instantly eliminate π΅ and πΆ. So we substitute π‘ equals zero
into our equation. And we find that one is equal to π΄
times zero minus one times zero plus two, which simplifies to one equals negative
two π΄. Dividing both sides of this
equation by negative two, and we obtain π΄ to be equal to negative one-half.

Next, we spot that if we let π‘ be
equal to one, weβre going to eliminate π΄ and πΆ. Substituting π‘ equals one gives us
one is equal to three π΅. And when we divide by three, we see
that π΅ is equal to one-third. Finally, weβll let π‘ be equal to
negative two. This will eliminate π΄ and π΅ this
time. This gives us one is equal to six
πΆ. And dividing through by six, we
find πΆ to be equal to one-sixth.

Letβs clear some space for the next
step. We can say that one over π‘ times
π‘ minus one times π‘ plus two is equal to negative one over two π‘ plus one over
three times π‘ minus one plus one over six times π‘ plus two. Weβre now ready to integrate with
respect to π‘. The integral of one over π‘ is the
natural log of the absolute value of π‘. So our first term integrates to
negative a half times the natural log of π‘. The interval of one over three
times π‘ minus one is a third times the natural log of the absolute value of π‘
minus one. And the integral of one over six
times π‘ plus two is equal to a sixth times the natural log of the absolute value of
π‘ plus two. We mustnβt forget since weβre
dealing with an indefinite integral to add that constant of integration πΆ.