Question

If $u_n=\int \frac{x^n}{\sqrt{a{x}^2+2bx+c}}dx$ then $a{u}_{n+1}+2b{u}_n+c{u}_{n-1}=\sqrt{a{x}^2+2bx+c}\frac{x^{n+k_1}}{n+k_1}-\frac{a{u}_{n+k_3}}{n+k_2}-\frac{b{u}_{n+k_4}}{n+k_2}$ then

Answer 1

$a{u}_{n+1}+2b{u}_n+c{u}_{n-1}=\sqrt{a{x}^2+2bx+c}\frac{x^{n+k_1}}{n+k_1}-\frac{a{u}_{n+k_3}}{n+k_2}-\frac{b{u}_{n+k_4}}{n+k_2}\cdots \text{(i)}$
Taking L.H.S.
$a{u}_{n+1}+2b{u}_n+c{u}_{n-1}=\int \frac{a{x}^{n+1}+2b{x}^n+c{x}^{n-1}}{\sqrt{a{x}^2+2bx+c}}dx$
$=\int \frac{x^{n-1}(a{x}^2+2bx+c)}{\sqrt{a{x}^2+2bx+c}}dx$
$=\int x^{n-1}\sqrt{a{x}^2+2bx+c}dx$
Let $v_n=\int x^{n-1}\sqrt{a{x}^2+2bx+c}dx$
$=\sqrt{a{x}^2+2bx+c}\frac{x^n}{n}-\frac{1}{n}\int \frac{x^n(ax+b)}{\sqrt{a{x}^2+2bx+c}}dx$
$=\sqrt{a{x}^2+2bx+c}\frac{x^n}{n}-\frac{a}{n}\int \frac{x^{n+1}}{\sqrt{a{x}^2+2bx+c}}dx-\frac{b}{n}\int \frac{x^n}{\sqrt{a{x}^2+2bx+c}}dx$
$\Rightarrow v_n=\sqrt{a{x}^2+2bx+c}\frac{x^n}{n}-\frac{a{u}_{n+1}}{n}-\frac{b{u}_n}{n}$
$\Rightarrow a{u}_{n+1}+2b{u}_n+c{u}_{n-1}=\sqrt{a{x}^2+2bx+c}\frac{x^n}{n}-\frac{a{u}_{n+1}}{n}-\frac{b{u}_n}{n}$
On comparing with equation (i)
$k_1=0,k_2=0,k_3=1,k_4=0$