better image integration

\hspace{-16}$For $\bf{(2)\;}$ Given $\bf{I=\int\frac{x^n}{\sqrt{ax^2+bx+c}}dx}$ and $\bf{n\in \mathbb{N}\;\;, a\neq 0}$\\\\\\ So $\bf{I(n+1) = \int\frac{ax^{n+1}}{\sqrt{x^2+bx+c}}dx = \frac{1}{2a}\int\frac{(2ax+b)x^n-bx^n}{\sqrt{ax^2+bx+c}}dx}$\\\\\\ So $\bf{I(n+1)=\frac{1}{2a}\left(\int\frac{(2ax+b)x^n}{\sqrt{ax^2+bx+c}}dx\right)-\frac{b}{2a}\int\frac{x^n}{\sqrt{ax^2+bx+c}}\right)dx}$\\\\\\ Now Using Integration by parts for $\bf{(1)}$ one....\\\\\\ So $\bf{I(n+1)=\frac{1}{2a}\left[2x^n\cdot \sqrt{ax^2+bx+c}-2n\int \sqrt{ax^2+bx+c}\cdot x^{n-1}\right]-\frac{b}{2a}I(n)}$\\\\\\ So $\bf{I(n+1)=\frac{x^n\sqrt{ax^2+bx+x}}{a}-\frac{n}{a}\int\frac{x^{n-1}\cdot \left(ax^2+bx+c\right)}{\sqrt{ax^2+bx+x}}dx-\frac{b}{2a}I(n)}$\\\\\\ So $\bf{I(n+1) = \frac{x^n\sqrt{ax^2+bx+x}}{a}-\frac{n}{a} \int \frac{ax^{n+1}+bx^n+cx^{n-1}}{\sqrt{ax^2+bx+c}}dx -\frac{b}{2a}I(n)}$\\\\\\ So $\bf{I(n+1) = \frac{x^n\sqrt{ax^2+bx+x}}{a}-nI(n+1)-\frac{b\cdot n}{a}I(n)-\frac{c\cdot n}{a}I(n-1)-\frac{b}{2a}I(n)}$\\\\\\

\hspace{-16}$So $\bf{(n+1)\cdot I(n+1)=\frac{x^n\cdot \sqrt{ax^2+bx+c}}{a}-\frac{b(2n+1)}{2a}I(n)-\frac{c\cdot n}{a}I(n-1)}$\\\\\\ So $\boxed{\boxed{\bf{I(n+1)=\frac{x^n\cdot \sqrt{ax^2+bx+c}}{a\cdot (n+1)}-\frac{b(2n+1)}{2a\cdot (n+1)}I(n)-\frac{c\cdot n}{a\cdot (n+1)}I(n-1)}}}$