calka
: 29 gru 2008, o 12:12
\(\displaystyle{ \int \frac{dx}{-5+6x-x ^{2} } = t \frac{dx}{-(x ^{2}-6x+5) } = t \frac{dx}{-[(x-1)(x-5)]} = - t \frac{dx}{(x-1)(x-5)}=...}\)
czy mozna tak to rozwiazywac?
\(\displaystyle{ \frac{dx}{x-1)(x-5)} = \frac{A}{x-1}+ \frac{B}{x-5}}\)
\(\displaystyle{ 1=A(x-5)+B(x-1)}\)
\(\displaystyle{ 1=(A=B)x-5A-B}\)
\(\displaystyle{ \begin{cases} A+B=0 \\ -5A-B=1 \end{cases}}\)
\(\displaystyle{ \begin{cases} A= -\frac{1}{4} \\ B= \frac{1}{4} \end{cases}}\)
\(\displaystyle{ ...= \frac{1}{4} t \frac{dx}{x-1}- \frac{1}{4} t \frac{dx}{x-5} = \frac{1}{4}ln|x-1| - \frac{1}{4}ln|x-5|+C}\)
to bedzie ok?
czy mozna tak to rozwiazywac?
\(\displaystyle{ \frac{dx}{x-1)(x-5)} = \frac{A}{x-1}+ \frac{B}{x-5}}\)
\(\displaystyle{ 1=A(x-5)+B(x-1)}\)
\(\displaystyle{ 1=(A=B)x-5A-B}\)
\(\displaystyle{ \begin{cases} A+B=0 \\ -5A-B=1 \end{cases}}\)
\(\displaystyle{ \begin{cases} A= -\frac{1}{4} \\ B= \frac{1}{4} \end{cases}}\)
\(\displaystyle{ ...= \frac{1}{4} t \frac{dx}{x-1}- \frac{1}{4} t \frac{dx}{x-5} = \frac{1}{4}ln|x-1| - \frac{1}{4}ln|x-5|+C}\)
to bedzie ok?