Question

If $U_n=\int \frac{x^n}{\sqrt{a{x}^2+2bx+c}}dx$, then $(n+1)a{u}_{n+1}+(2n+1)b{u}_n+nc{u}_{n-1}$ is equal to :

• A. $x^{n-1}\sqrt{a{x}^2+2bx+c}$
• B. $\frac{x^{n-2}}{\sqrt{a{x}^2+2bx+c}}$
• C. $\frac{x^n}{\sqrt{a{x}^2+2bx+c}}$
• D. $x^n\sqrt{a{x}^2+2bx+c}$

Answer 1

The correct option is D $x^n\sqrt{a{x}^2+2bx+c}$

$U_{n-1}=\int \frac{x^{n-1}}{\sqrt{a{x}^2+2bx+c}}dx$
$=\frac{1}{2a}\int \frac{x^n(2ax+2b)-2b{x}^n}{\sqrt{a{x}^2+2bx+c}}dx$
$=\frac{1}{2a}\int \frac{x^n(2ax+2b)}{\sqrt{a{x}^2+2bx+c}}dx-\frac{b}{a}{U}_n$
$=I_n-\frac{b}{a}{U}_n$ , where ....(i)
$I_n=\frac{1}{2a}\int \frac{x^n(2ax+2b)}{\sqrt{a{x}^2+2bx+c}}dx$
$=\frac{1}{2a}\int \frac{x^n(2ax+2b)}{a{x}^2+2bx+c}dx$
$=\frac{1}{2a}{x}^{n}2\sqrt{a{x}^2+2bx+c}$
$-\int n{x}^{n-1}2\sqrt{a{x}^2+2bx+c}dx$
$=\frac{x^n}{a}\sqrt{a{x}^2+2bx+c}dx$ .... (ii)
from (i) and (ii) be get
$(n+1)a{u}_{n+1}+(2n+1)b{u}_n+nc{u}_{n-1}$ $=x^n\sqrt{a{x}^2+2bx+c}$