<s>[INST] If $I_m=\int \frac{x^m}{\sqrt{ax^2+c}}\, dx$, show that $amI_m=x^{m-1}\sqrt{ax^2+c}-(m-1)c I_{m-2}$. Problem: If $I_m=\int \dfrac{x^m}{\sqrt{ax^2+c}}\, dx$,  show that  $amI_m=x^{m-1}\sqrt{ax^2+c}-(m-1)c I_{m-2}$.  
Observe
\begin{align*}
 I_m= & \int \dfrac{x^m}{\sqrt{ax^2+c}}\, dx\\
 = &  \dfrac{1}{a}\int x^{m-1}d(\sqrt{ax^2+c}) \\
 = & \dfrac{1}{a}\left[\dfrac{x^{m-1}}{\sqrt{ax^2+c}}-\int (m-1)x^{m-2} d( \sqrt{ax^2+c})\right] \\
 =& \dfrac{1}{a}\left[x^{m-1}\sqrt{ax^2+c}-(m-1)I_{m-2}\right]
\end{align*}
This implies  $aI_m=x^{m-1}\sqrt{ax^2+c}-(m-1)I_{m-2}$ which is different from required result. Which one is correct?
 [/INST] Your third and fourth steps are wrong. The correct steps are 
$$I_m = \frac{1}{a}\Bigg[x^{m-1}\sqrt{ax^2+c}-\int(m-1)x^{m-2}\sqrt{ax^2+c}\ dx\Bigg]$$
$$=\frac{1}{a}\Bigg[x^{m-1}\sqrt{ax^2+c}-(m-1)\int \frac{x^{m-2}(ax^2+c)}{\sqrt{ax^2+c}}dx\Bigg]$$
$$=\frac{1}{a}\Bigg[x^{m-1}\sqrt{ax^2+c}-a(m-1)I_m-c(m-1)I_{m-2}\Bigg]$$
and you get the desired result in next step.
Hope it helps:)
</s>