So I was bored, and decided to evaluate some integrals. After a while, I came up with this:$$\int_{-100}^{+100}\int_{-100}^{+100}(2y+1)e^{2y\ln(x)}dxdy$$which I thought that I might be able to evaluate. Here is my attempt at doing so:$$\int_{-100}^{+100}\int_{-100}^{+100}(2y+1)e^{2y\ln(x)}dxdy\implies\int_{-100}^{+100}\int_{-100}^{+100}(2y+1)x^{2y}dxdy$$$$\implies\left.\dfrac{x^{2y+1}}{\require{cancel}\cancel{2y+1}}\right]_{-100}^{+100}\cancel{(2y+1)}\implies100\int_{-100}^{+100}10000^y+10000^ydy$$$$=100\int_{-100}^{+100}20000^ydy\implies100\left[\dfrac{20000^y}{\ln(20000)}\right]_{-100}^{+100}=100\left(\dfrac{20000^{200}-1}{20000^{100}\ln(20000)}\right)$$$$=\dfrac{4^{100}100^{401}-100}{20000^{100}\ln(20000)}$$$$\approx1.28000423437055155\cdot10^{431}$$as seen by this result by Wolfram Alpha, so this is the result from the evaluation is as seen above.
My question
Is my evaluation of the integral correct, or what could I do to evaluate it/evaluate it more quickly?