# Evaluate the following iterated integrals. int_0^2int_0^1frac{8xy}{1+x^4}dxdy

Evaluate the following iterated integrals.
${\int }_{0}^{2}{\int }_{0}^{1}\frac{8xy}{1+{x}^{4}}dxdy$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

StrycharzT

$\text{Evaluate the given integral as shown below:}$
${\int }_{0}^{2}{\int }_{0}^{1}\frac{8xy}{1+{x}^{4}}={\int }_{0}^{2}y\left[{\int }_{0}^{1}\frac{8x}{1+{x}^{4}}dx\right]dy$

${\int }_{0}^{2}{\int }_{0}^{1}\frac{8xy}{1+{x}^{4}}dxdy={\int }_{0}^{2}y\left[{\int }_{0}^{1}\frac{8x}{1+{x}^{4}}dx\right]dy$
$={\int }_{0}^{2}y\left[4{\int }_{0}^{1}\frac{2x}{1+\left({x}^{2}{\right)}^{2}}dx\right]dy$
$={\int }_{0}^{2}y\left[4{\int }_{0}^{1}\frac{du}{1+{u}^{2}}dx\right]dy$
$={\int }_{0}^{2}y\left[4\left({\mathrm{tan}}^{-1}u{\right)}_{0}^{1}\right]dy$
$={\int }_{0}^{2}y\left[4\left({\mathrm{tan}}^{-1}1-{\mathrm{tan}}^{-1}0\right)\right]dy$
$={\int }_{0}^{2}y\left[4\left(\pi /4-0\right)\right]dy$
$={\int }_{0}^{2}y\pi dy$
$=\pi \left[\frac{{y}^{2}}{2}{\right]}_{0}^{2}$
$=\pi \left[\frac{{2}^{2}}{2}-0{\right]}_{0}^{2}$
$=2\pi$