Question

∫20x(5x²+x-2) dx

∫4(10^{2x^2}) dx

Answer 1

$\int(100x^3+20x^2-40x)dx=25x^4+20x^3/3-20x^3+C$

$4\smallint10^{2x^2}dx$

$u=\sqrt{2}x$

$du=\sqrt2dx$

$4\smallint10^{2x^2}dx=\frac{4}{\sqrt2}\smallint10^{u^2}du=4\frac{\sqrt{π}}{ 2^{3/2}\sqrt{ln(10)}}\intop\frac{2\sqrt{ln(10)}10^{u^2}}{\sqrtπ}du$

$\intop\frac{2\sqrt{ln(10)}10^{u^2}}{\sqrtπ}du=erfi(\sqrt{ln(10)}u)$

$4\smallint10^{2x^2}dx=4\frac{\sqrt{π}}{ 2^{3/2}\sqrt{ln(10)}}erfi(\sqrt{ln(10)}u)=4\frac{\sqrt{π}}{ 2^{3/2}\sqrt{ln(10)}}erfi(\sqrt{ln(10)}\sqrt2x)+C$