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Below is a problem from the book "Calculus and Analytic Geometry" by Thomas Finney. I am hoping somebody can check my work. I consider it to be a particular hard problem.
Thanks,

Bob

Problem:

We dare you to evalaute this integral.
$$ int frac1x(x+1)(x+2)(x+3)(x+4) ... (x+m) ,, dx $$
Answer:

To evaluate this integral, we will consider some special cases. For $m = 0$ we have:
beginalign*
int frac1x ,, dx &= ln|x| + C \
endalign*
Now for $m = 1$ we have the following integral:
$$ int frac1x(x+1) ,, dx $$
beginalign*
frac1x(x+1) &= fracAx + fracBx+1 \
1 &= A(x+1) + B(x) \
endalign*
At $x = 0$ we have $1 = A(0+1)$ which yields $A = 1$.
beginalign*
A + B &= 0 \
1 + B &= 0 \
B &= -1 \
frac1x(x+1) &= frac1x - frac1x+1 \
int frac1x(x+1) ,, dx &= ln|x| - ln|x+1| + C \
endalign*
Now for $m = 2$ we have the following integral:
$$ int frac1x(x+1)(x+2) ,, dx $$
beginalign*
frac1x(x+1)(x+2) &= fracAx + fracBx+1 + fracCx+2 \
1 &= A(x+1)(x+2) + B(x)(x+2) + C(x)(x+1) \
endalign*
At $x = 0$ we have $1 = A(0+1)(0+2)$ which yields $A = frac12$.
newline
At $x = -1$ we have $1 = B(-1)(-1+2) = -B$ which yields $B = -1$.
newline
At $x = -2$ we have $1 = C(-2)(-2+1) = 2C$ which yields $C = frac12$.
beginalign*
frac1x(x+1)(x+2) &= frac frac12x - frac1x+1 + frac frac12x+2 \
int frac1x(x+1)(x+2) ,, dx &= frac12 ln - ln + frac12 ln + C \
endalign*
newline
Now for $m = 3$ we have the following integral:
$$ int frac1x(x+1)(x+2)(x+3) ,, dx $$
beginalign*
frac1x(x+1)(x+2)(x+3) &= fracAx + fracBx+1 + fracCx+2 + fracDx+3 \
1 &= A(x+1)(x+2)(x+3) + B(x)(x+2)(x+3) + \
& C(x)(x+1)(x+3) + D(x)(x+1)(x+2) \
endalign*
At $x = 0$ we have $1 = A(0+1)(0+2)(0+3) = 6A$ which yields $A = frac16$.

At $x = -1$ we have $1 = B(-1)(-1+2)(-1+3) = -2B$ which yields $B = -frac12$.

At $x = -2$ we have $1 = C(-2)(-2+1)(-2+3) = 2C$ which yields $C = frac12$.

At $x = -3$ we have $1 = D(-3)(-3+1)(-3+2) = -6D$ which yields $D = -frac16$.

Hence, we have the following solution:
$$ int frac1x(x+1)(x+2)(x+3) ,, dx =
frac16ln - frac12ln + frac12ln - frac16ln + C $$
newline
Now for $m = 4$ we have the following integral:
$$ int frac1x(x+1)(x+2)(x+3)(x+4) ,, dx $$
beginalign*
frac1x(x+1)(x+2)(x+3)(x+4) &= fracAx + fracBx+1 + fracCx+2 + fracDx+3 + fracEx+4 \
1 &= A(x+1)(x+2)(x+3)(x+4) + B(x)(x+2)(x+3)(x+4) + \
&+ C(x)(x+1)(x+3)(x+4) \
&+ D(x)(x+1)(x+2)(x+4) + E(x)(x+1)(x+2)(x+3) \
endalign*
At $x = 0$ we have $1 = A(0+1)(0+2)(0+3)(0+4) = 24A$ which yields $A = frac124$.

At $x = -1$ we have $1 = B(-1)(-1+2)(-1+3)(-1+4) = -6B$ which yields $B = -frac16$.

At $x = -2$ we have $1 = C(-2)(-2+1)(-2+3)(-2+4) = 4C$ which yields $C = frac14$.

At $x = -3$ we have $1 = D(-3)(-3+1)(-3+2)(-3+4) = -6D$ which yields $D = -frac16$.

At $x = -4$ we have $1 = E(-4)(-4+1)(-4+2)(-4+3) = 24E$ which yields $E = frac124$.

Hence we have the solution:

Now for $m = 4$ we have the following integral:
$$ int frac1x(x+1)(x+2)(x+3)(x+4) ,, dx =
fracln - 4ln + 6ln - 4ln + ln 24 + C $$
Now let's consider the general case.
beginalign*
frac1x(x+1)(x+2)(x+3)(x+4) cdots (x+m) ,, &= fracC_0x + fracC_1x+1 cdots + fracC_mx+m \
endalign*
beginalign*
1 &= C_0(x+1)(x+2) cdots (x+m) + C_1(x)(x+2) cdots (x+m) + C_2(x)(x+1)(x+3)(x+4) cdots (x+m) + \
& cdots C_m(x)(x+1)(x+2) cdots (x+m-1) \
endalign*
Now lets consider the first term. We set $x = 0$ and we get:
beginalign*
1 &= C_0(0+1)(0+2) cdots (x+m) = m! C_0 \
C_0 &= frac1m!
endalign*
Now lets consider the $C_2$ term. We set $x = 2$ and $m > 4$. We get:
beginalign*
1 &= C_2(-2)(-2+1)(-2+3)(-2+4)(-2 + 5) cdots (-2 + m) \
1 &= C_2 (2)(1)(2)(3)(4) cdots (m-2) \
1 &= 2(m-2)! C_2 \
C_2 &= frac12(m-2)! = fracm(m-1)2(m!) \
C_2 &= frac binom m2 m! \
endalign*
Now lets consider the last term. We set $x = -m$ and we get:
beginalign*
1 &= C_m(-m)(-m+1)(-m+2) cdots (x+m -1) \
C_m &= frac(-1)^mm! \
endalign*
Now lets consider one of the middle terms. We set $x = -k$ where $0 <= k <= m$ and we get:
beginalign*
1 &= C_k(-k)(-k+1)(-k+2) cdots (-1) (1)(2) cdots (-k + m - 1) \
1 &= -1^k C_k(k-1)(k-2) cdots (1)(2) cdots (-k + m - 1) \
C_k &= frac 1 (-1)^k C_k(k-1)(k-2) cdots (1)(2) cdots (-k + m - 1) \
C_k &= frac k! (-1)^k C_k(k-1)(k-2) cdots (1)m! \
C_k &= frac binom mk (-1)^k m! \
endalign*
Hence the answer is:
$$ sum_k=0^k=m left( frac binom mk (-1)^k m! right)
lnx+k + C $$

At this line: "Now lets consider the first term. We set $x=0$ and we get:"
$$1 = C_0(0+1)(0+2) cdots (x+m) = m!C_0$$
When you set $x=0$ there won't be any $x$ left.

For $C_2$ there is no mistake but then for the general term you did the same typo.

You can't have after you set $x=-m$: $$1 = C_m(-m)(-m+1)(-m+2) cdots (colorredx+m -1)$$
And for the last part for some reason, you have $C_k$ in the denominator.
$$beginalign*
1 &= C_k(-k)(-k+1)(-k+2) cdots (-1) (1)(2) cdots (-k + m - 1) \
1 &= (-1)^k C_k(k-1)(k-2) cdots (1)(2) cdots (-k + m - 1) \
C_k &= frac 1 (-1)^k colorredC_k(k-1)(k-2) cdots (1)(2) cdots (-k + m - 1) \
C_k &= frac k! (-1)^k colorredC_k(k-1)(k-2) cdots (1)m! \
C_k &= frac binom mk (-1)^k m! \
endalign*$$
Other than this everything it's correct.

A simpler approach is possible without all the other preliminary work. Let $$q_m(x) = prod_k=0^m (x+k), quad f_m(x) = frac1q_m(x).$$ Then $f$ admits a partial fraction decomposition of the form $$f_m(x) = sum_n=0^m fracA_nx+n tag1$$ for suitable constants $A_0, ldots, A_m$ which we wish to find, hence an antiderivative of $f$ is $$int f_m(x) , dx = sum_n=0^m A_n log |x+n|. tag2$$ (I have omitted the constant of integration for convenience.) So all that remains is to determine the form of $A_n$. To do this, we observe that $$1 = q_m(x) sum_n=0^m fracA_nx+n = sum_n=0^m p_n(x) A_n,$$ where $$p_n(x) = prod_k ne n (x+k) = (-1)^n prod_k=0^n-1 (-x-k) prod_k=n+1^m (k + x).$$ Then in particular $$p_n(-n) = (-1)^n prod_k=0^n-1 (n-k) prod_k=n+1^m (k-n) = (-1)^n n!(m-n)! = frac(-1)^n m!binommn,$$ and $p_n(-k) = 0$ for all other nonnegative integers $k le m$ not equal to $n$. Therefore, $$A_n = frac1p_n(-n) = frac(-1)^nm! binommn$$ and $$int f_m(x) , dx = sum_n=0^m frac(-1)^nm! binommn log |x+n| + C$$ as claimed.

It looks right to me [That is, I did the calculation independently and got the same answer]. For brevity/clarity purposes you really only need to include the “general case”.

It would be more conventional to write
$$
C_k = frac(-1)^kk!(m-k)!
$$

We could set up a difference equation. Partial fractions indicates that
$$f_m(x)=frac1prod_k=0^m(x+k)=sum_k=0^mfracA_k^(m)x+m$$
And it is a simple calculation to show that
$$f_m-1(x)-f_m-1(x+1)=mf_m(x)$$
So, comparing coefficients of $frac1x+k$ we have $A_0^(m-1)=mA_0^(m)$, $A_m-1^(m-1)=-mA_m^(m)$, and $A_k^(m-1)-A_k-1^(m-1)=mA_k^(m)$ for $1le kle m-1$. If we let $A_k^(m)=frac(-1)^km!B_k^(m)$ then our difference equations read $B_0^(m-1)=B_0^(m)=cdots=B_0^(0)=A_0^(0)=1$, $B_m-1^(m-1)=B_m^(m)=cdots=B_0^(0)=1$ and $B_k^(m-1)+B_k-1^(m-1)=B_k^(m)$ for $1le kle m-1$. We recognize these as the difference equations for Pascal's triangle, so $B_k^(m)=mchoose k$ so it follows that $A_k^(m)=frac(-1)^km!mchoose k$ and
$$int f_m(x)dx=sum_k=0^mfrac(-1)^km!mchoose kln|x+k|+C$$