Question

Solve the following quadratic equations $\left(i\right)8x^2+2x+1=0$

Answer 1

$\int x\ln \sqrt{x}dx=\frac{1}{2}\int x\ln xdx$

$=\frac{1}{2}[\frac{x^2}{2}\ln x-\int dx\frac{x}{2}\Rightarrow \frac{x^2}{4}\ln x-\frac{x^2}{8}+c$

$8x^2+2x+1=0$

$x=\frac{-2\pm \sqrt{4-4\left(8\right)\left(1\right)}}{2}=\frac{-2\pm \sqrt{8i}}{2}$

$I=\int 5^{5^{5^x}}.5^{5^x}.5^{x}dx$

$5^{5^{5^x}}=t$

$5^{5^{5^5}}.5^{5^x}.5^x(\log 5{)}^{3}dx=dt$

$I=\frac{1}{(\log 5{)}^3}\int dt=\frac{t}{(\log 5{)}^3}+c$

$=\frac{5^{5^{5^x}}}{(\log 5{)}^3}+c$