I am studying the Oxford Concise Dictionary of Mathematics by Richard Earl and James Nicholson, 6th Edition (2021). There are some new results in Appendix 8: Integrals which were added since the 5th edition. At least two of these are demonstrably wrong, and this next one I've been trying to resolve that I think may also be wrong:
For the integral
$$I_{n, k} = \int x^k \left({x^2 + A x + B}\right)^n \ \mathrm d x$$
with $n \ge 0$, $k \ge 2$, show the reduction formula
$$I_{n, k} = \dfrac {x^{k-1} \left({x^2 + A x + B}\right)^{n + 1} }{\left({k + 2 n + 1}\right)} – \dfrac {B \left({k – 1}\right)} {\left({k + 2 n + 1}\right)} I_{n, k-2} – \dfrac {A \left({k + n}\right)} {\left({k + 2 n + 1}\right)} I_{n, k-1}$$
My approach is to use integration by parts, using $\int u \ \mathrm d v = u v – \int v \ \mathrm d u$ where $u = \left({x^2 + A x + B}\right)^n$ and $\mathrm d v = x^k$, but from the form of the solution it appears as though it begs for $u$ and $\mathrm d v$ to be the other way round (as in the $u v$ part the power of $u$ has decreased and the power of $v$ has increased).
But using $\mathrm d v = \left({x^2 + A x + B}\right)^n$ gives a difficult integration to find $v$ which is almost as troublesome as this beast we're trying to solve.
Hence I tried $u = \left({x^2 + A x + B}\right)^n$ and $\mathrm d v = x^k$, which are more manageable.
This gives $\mathrm d u = n \left({2 x + A}\right) \left({x^2 + A x + B}\right)^{n – 1}$ and $v = \frac {x^{k + 1} } {k + 1}$.
Flogging through the algebra gets us to:
$$I_{n, k} = \dfrac {x^{k + 1} \left({x^2 + A x + B}\right)^n} {k + 1} – \dfrac {2 n} {k + 1} I_{n – 1, k + 2} – \dfrac {A n} {k + 1} I_{n – 1, k + 1}$$
which is a good start, and we can get to the required powers in the fraction by using the above to evaluate instead:
$$I_{n + 1, k – 2} = \dfrac {x^{k – 1} \left({x^2 + A x + B}\right)^{n + 1} } {k – 1} – \dfrac {2 \left({n + 1}\right) } {k – 1} I_{n, k} – \dfrac {A \left({n + 1}\right) } {k – 1} I_{n, k – 1}$$
Rearranging this, we get:
$$I_{n, k} = \dfrac {x^{k – 1} \left({x^2 + A x + B}\right)^{n + 1} } {2 n + 2} – \dfrac {k – 1} {2 n + 2} I_{n + 1, k – 2} – \dfrac {A \left({n + 1}\right) } {2 n + 2} I_{n, k – 1}$$
Close, but nowhere near a cigar.
Anyone able to offer a hint as to how to proceed, or what I may have done wrong?
Best Answer
Let $h(x)=x^2+Ax+B$ and differentiate
\begin{align} & \frac d{dx}\left[x^{k-1}h^{n+1}(x)\right]\\ = &\left[ (k+2n+1)x^k +A(k+n)x^{k-1}+B(k-1) x^{k-2}\right]h^n(x) \end{align}
Integrate both sides and then rearrange to obtain the sought-after reduction formula
\begin{align} I_{n, k} =&\int x^k (x^2+Ax+B )^n dx \\ =& \ \frac1{k+2n+1}\left[ x^{k-1} h^{n + 1} (x) - {A \left({k + n}\right)} I_{n, k-1}- {B \left({k - 1}\right)} I_{n, k-2}\right] \end{align}