I am trying this for the last few days but no way out. can anybody show me some light on this integral?
$\int{ (Mx + N) \over(A +2Bx+Cx^2)^{p}}dx={(NB-MA)+(NC-MB)x \over(2(p-1)(AC-B^2)(A+2Bx+Cx^2)^{p-1}}+{(2p-3)(NC-MB) \over 2(p-1)(AC-B^2)} \int {dx \over (A+2Bx+Cx^2)^{p-1} } $
Right, lets start with typo corrections \begin{eqnarray*} \int \frac{Mx+N}{(A+2Bx+Cx^2)^{\color{red}{p}}} = \frac{(NB-MA)+(NC-MB)x}{2(p-1)(AC-B^2)(A+2Bx+Cx^2)^{p-1}}+ \end{eqnarray*} \begin{eqnarray*}\frac{\color{red}{(2p-3)}(NC-MB)}{2(p-1)(AC-B^2)} \int\frac{dx}{(A+2Bx+Cx^2)^{p-1}} \end{eqnarray*} We shall show this, by differentiating the RHS and showing that it gives the integrand of the LHS. First note that
\begin{eqnarray*} \frac{d}{dx} \frac{1}{(A+2Bx+Cx^2)^{p-1}} = \frac{-2(p-1)(B+Cx)}{(A+2Bx+Cx^2)^{p}}. \end{eqnarray*} Differentiating the RHS gives \begin{eqnarray*} \frac{-(B+Cx)((NB-MA)+(NC-MB)x}{(AC-B^2)(A+2Bx+Cx^2)^{p}}+\frac{(NC-MB)}{2(p-1)(AC-B^2)(A+2Bx+Cx^2)^{p-1}}+\frac{(NC-MB)(2p-3)}{2(p-1)(AC-B^2)(A+2Bx+Cx^2)^{p-1}} \end{eqnarray*} add the second & third terms gives \begin{eqnarray*} \frac{-(B+Cx)((NB-MA)+(NC-MB)x)}{(AC-B^2)(A+2Bx+Cx^2)^{p}}+\frac{(NC-MB)(1+2p-3)}{2(p-1)(AC-B^2)(A+2Bx+Cx^2)^{p-1}} \end{eqnarray*} cancal $2(p-1)$ and multiply the top by $(A+2Bx+Cx^2)$ and add these terms gives \begin{eqnarray*} \frac{-(B+Cx)((NB-MA)+(NC-MB)x)+(NC-MB)(A+2Bx+Cx^2)}{(AC-B^2)(A+2Bx+Cx^2)^{p}} \end{eqnarray*} After a little algebra in the numerator & cancelling a the factor $AC-B^2$ top and bottom we have \begin{eqnarray*} \frac{Mx+N}{(AC-B^2)(A+2Bx+Cx^2)^{p}} \end{eqnarray*} precisely the integrand of the LHS, as required.