bounded by y=x^2, y=4, x=0. picture: graph y = x^2, y=4, and x=0 from x=0 to x=2 - Wolfram|Alpha
i set my integral: integrate 0 to 2 [ integrate x^2 to 4 x(1+y^2)^-0.5 dy ] dx
then I tried doing it, first integral results: x[arcsinh(y)] from 4 to x^2
but second integral, how to integrate x*arcsinh(x^2).
edit: is this how
[[x^2⋅arcsinh(x^2)] - [(x^4 +1)^.5]]/2
Horizontal strips are a better choice than vertical in this case. So $\displaystyle \begin{align*} 0 \leq x \leq \sqrt{y} \end{align*}$ and $\displaystyle \begin{align*} 0 \leq y \leq 4 \end{align*}$, so that means your double integral is
$\displaystyle \begin{align*} \int_0^4{\int_0^{\sqrt{y}}{x \left( 1 + y^2 \right) ^{-\frac{1}{2}} \,dx}\,dy} &= \int_0^4{ \left[ \frac{x^2}{2} \left( 1 + y^2 \right) ^{-\frac{1}{2}} \right]_0^{\sqrt{y}} \,dy } \\ &= \int_0^4{ \left[ \frac{ \left( \sqrt{y} \right) ^2 }{2} \left( 1 + y^2 \right) ^{-\frac{1}{2}} \right] - \left[ \frac{0^2}{2} \left( 1 + y^2 \right) ^{-\frac{1}{2}} \right] \,dy } \\ &= \frac{1}{2}\int_0^4{ y \, \left( 1 + y^2 \right) ^{-\frac{1}{2}} \, dy }\\ &= \frac{1}{4} \int_0^4{ 2y \, \left( 1 + y^2 \right) ^{-\frac{1}{2}} \, dy } \\ &= \frac{1}{4} \int_1^{17}{ u^{-\frac{1}{2}} \, du } \textrm{ when we substitute } u =1 + y^2 \\ &= \frac{1}{4} \left[ 2u^{\frac{1}{2}} \right]_1^{17} \\ &= \frac{1}{4} \left[ 2 \left( 17 \right) ^{\frac{1}{2}} - 2 \left( 1 \right) ^{\frac{1}{2}} \right] \\ &= \frac{1}{2} \left( \sqrt{17} - 1 \right) \end{align*}$
ahh, i considered doing the horizontal method but I didn't see how it would alleviate the difficult integral but seeing it written out like this, you have Proven It!!
I am adding extra smilies to my sig to show my appreciation
Reversing the order of integration, as both Idea and Prove It said is the best way to handle this integral. But to answer you last question, "how to integrate x arcsinh(x^2)" let u= x^2 so [tex]\int x arcsinh(x) dx[tex] becomes and that can be looked up in a table of integrals.
(Lists of integrals - Wikipedia, the free encyclopedia)
Or otherwise derived (note I've left out the integration constants until the end)...
Let's consider $\displaystyle \begin{align*} \int{ \frac{2x}{\sqrt{1 + x^2}} \, dx} = \int{ 2x \left( 1 + x^2 \right) ^{-\frac{1}{2}} \,dx} \end{align*}$. With the substitution $\displaystyle \begin{align*} u = 1 + x^2 \implies du = 2x\,dx \end{align*}$ we can see $\displaystyle \begin{align*} \int{ 2x \left( 1 + x^2 \right) ^{-\frac{1}{2}} \, dx} = \int{ u^{-\frac{1}{2}} \, du } = 2u^{\frac{1}{2}} = 2 \, \sqrt{ 1 + x^2 } \end{align*}$.
We might also consider integration by parts:
$\displaystyle \begin{align*} \int{ \frac{2x}{\sqrt{1 + x^2}} \, dx} &= \int{ 2x \, \frac{1}{\sqrt{1 + x^2}} \, dx} \\ &= 2x\,\textrm{arsinh}\,{(x)} - \int{ 2\,\textrm{arsinh}\,{(x)}\,dx } \\ &= 2x\,\textrm{arsinh}\,{(x)} - 2\int{ \textrm{arsinh}\,{(x)}\,dx } \end{align*}$
and thus
$\displaystyle \begin{align*} 2x\,\textrm{arsinh}\,{(x)} - 2\int{ \textrm{arsinh}\,{(x)}\,dx} &= 2\,\sqrt{1 + x^2} \\ x\,\textrm{arsinh}\,{(x)} - \int{ \textrm{arsinh}\,{(x)}\,dx} &= \sqrt{1 + x^2} \\ \int{\textrm{arsinh}\,{(x)}\,dx} &= x\,\textrm{arsinh}\,{(x)} - \sqrt{1 + x^2} + C \end{align*}$