\(f(x)=\frac{1}{0,5(5x-1)^2} =\frac{2}{(5x-1)^2}\)
\(F(x)=2\cdot\int\limits_{}^{}\frac{1}{(5x-1)^2}dx\)
Substitution
\(5x-1=u\) → \(x=\frac{1}{5}u+\frac{1}{5}\) \( \frac{dx}{du}=\frac{1}{5} \) \( dx=\frac{1}{5}du \)
Einschub:
\(\int\limits_{}^{}\frac{1}{u^2} \cdot\frac{1}{5}du=\frac{1}{5}\cdot\int\limits_{}^{}\frac{1}{u^2} du=\frac{1}{5}\cdot\int\limits_{}^{}u^{-2}du \)
Einschub:
\(\int\limits_{}^{}u^{-2}du =\frac{ u^{-2+1}}{-2+1}=\frac{ u^{-1}}{-1}=- u^{-1}\)