calki pomoc, wyciaganie stalych przed calke

[mikolajkapica]

Dlaczego to:

\(\int{2\over{7x+14}}dx=\int{2\over7}*{1\over{x+2}}dx={2\over7}\int{dt\over{t}}={2\over7}ln|t|+C={2\over7}ln|x+2|+C\\
t = x + 2 \\
dt = dx \\\)

!=


\(\int{2\over{7x+14}}dx=\int{2{dt}\over{7t}}={{2}\over{7}}\int{{dt}\over{t}}={{2}\over{7}}ln|t| + C={{2}\over{7}}ln|7x+14| + C \\
t = 7x + 14 \\
dt = 7dx \\
{dt\over{7}}=dx\)

[jjjjjj]

Dlaczego to:

\(\int{2\over{7x+14}}dx=\int{2\over7}*{1\over{x+2}}dx={2\over7}\int{dt\over{t}}={2\over7}ln|t|+C={2\over7}ln|x+2|+C\\
t = x + 2 \\
dt = dx \\\)

!=


\(\int{2\over{7x+14}}dx=\int{2{dt}\over{7t}}={{2}\over{7}}\int{{dt}\over{t}}={{2}\over{7}}ln|t| + C={{2}\over{7}}ln|7x+14| + C \\
t = 7x + 14 \\
dt = 7dx \\
{dt\over{7}}=dx\)

\(
{{2}\over{7}}ln|7x+14| = {{2}\over{7}}ln(7|x+2|)={{2}\over{7}}(ln|x+2|+ln7)={{2}\over{7}}ln|x+2|+{{2}\over{7}}ln7\)
Potraktuj \( {{2}\over{7}}ln7\) jako stałą C - wszystko jest w porządku

[mikolajkapica]

Dziękuję!