Integrate a function with peculiar form I was reading through some problems on integration, I came across the following, 
Given,
$$
I_n = \int_{1}^{e^2}(\ln(x))^{n}d(x^{2})
$$
Then the value of $2I_n+nI_{n-1}$ is?
I tried substituting $x^{2}$ = $t$ and then doing it; but how do I change the limits? 
Any kind of help will be appreciated.
 A: No need to do substitution, use integration by parts directly:
$$I_n = \int_{1}^{e^2}(\ln(x))^{n}d(x^{2})$$
$$I_n=(\ln x)^nx^2 \big|_1^{e^2} - \int_1^{e^2}x^2n(\ln x)^{n-1}\frac{1}{x}dx$$ 
$$=(\ln x)^nx^2 \big|_1^{e^2} - \frac{n}{2}\int_1^{e^2}(\ln x)^{n-1}dx^2$$
$$=(\ln x)^nx^2 \big|_1^{e^2} - \frac{n}{2}I_{n-1}$$ 
Thus
$$2I_n + nI_{n-1}=2(\ln x)^nx^2\big|_1^{e^2} -nI_{n-1} + nI_{n-1}$$
$$=2\times2^n\times e^4 - 0 = e^4\times 2^{n+1}$$
A: \begin{align*}
I_n&=\int_{1}^{e^2}(\ln x)^n\ d(x^2)\\
\Rightarrow I_n&=2\int_{1}^{e^2}x(\ln x)^n\ dx\\
&=2\left[(\ln x)^n\cdot\int_{1}^{e^2}x\ dx-\int_{1}^{e^2}\left\{\dfrac{d}{dx}(\ln x)^n\cdot\left(\int_{1}^{e^2}x\ dx\right)\ dx\right\}\right]\\
&=2\left[\left.(\ln x)^n\cdot \dfrac{x^2}{2}\right|_{1}^{e^2}-\int_{1}^{e^2}n(\ln x)^{n-1}\dfrac{1}{x}\cdot\dfrac{x^2}{2}\ dx\right]\\
&=\left.(\ln x)^n\cdot x^2\right|_{1}^{e^2}-n\int_{1}^{e^2}x(\ln x)^{n-1}\ dx\\
&=2^n\cdot e^4-n\int_{1}^{e^2}x(\ln x)^{n-1}\ dx\\
\Rightarrow 2I_n&=2^{n+1}\cdot e^4-n\cdot2\int_{1}^{e^2}x(\ln x)^{n-1}\ dx\\
\Rightarrow 2I_n&=2^{n+1}\cdot e^4-nI_{n-1}\\
\Rightarrow 2I_n+nI_{n-1}&=2^{n+1}\cdot e^4
\end{align*}
