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\begin{document}

{\bf Question} if $I_n=\ds\int (x^2+3)^n \,dx$, show that
$(2n+1)I_n=x(x^2+3)^n+6nI_{n-1}$.


\medskip

{\bf Answer} $I_n=\ds\int(x^2+3)^n \,dx$

We want to reduce the powers of $n$. So, integrate by parts with

$$\begin{array} {cc} u=(x^2+3)^n & \ds\frac{dv}{dx}=1\\
\ds\frac{du}{dx}=n(x^2+3)^{n-1}2x & v=x \end{array}$$

$\begin{array} {rcl} I_n & = & x(x^2+3)-\ds\int\,dx \cdot x \cdot
n \cdot 2x \cdot (x^2+3)^{n-1}\\ & = & x(x^2+3)-2n\ds\int \,dx
x^2(x^2+3)^{n-1}\end{array}$

Now $x^2=(x^2+3)-3$. This is the key step

So

$\begin{array} {rcl} I_n & = & x(x^2+3)-2n\ds\int \,dx
[(x^2+3)-3](x^2+30^{n-1}\\ & = & x(x^2+3)-2n \underbrace{\ds\int
\,dx(x^2+3)^n}+6n\underbrace{\ds\int
\,dx(x^2+3)^{n-1}}\end{array}$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \  $I_n$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
$I_{n-1}$

So $I_n=x(x^2+3)-2nI_n+6nI_{n-1}$

Rearrange for $I_n$

\un{$(2n+1)I_n=x(x^2+3)+6nI_{n-1}$} as required.

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