całka

[malwik91]

obliczyc: \(\int \frac{x}{(x^2+1^2} dx\)

[domino21]

\(\int \frac{x}{(x^2+1)^2} dx =\left(x^2+1 =t \\ 2x dx =dt \\ x dx =\frac{1}{2} dt \right)=\frac{1}{2} \int \frac{dt}{t^2} =-\frac{1}{2} t^{-1} +C=-\frac{1}{2t} +C=-\frac{1}{2(x^2+1)}+C\)

[malwik91]

\(\int \frac{x}{(x^2+1)^2} dx =\left(x^2+1 =t \\ 2x dx =dt \\ x dx =\frac{1}{2} dt \right)=\frac{1}{2} \int \frac{dt}{t^2} =\frac{1}{2} t^{-1} +C=\frac{1}{2t} +C=\frac{1}{2(x^2+1)}+C\)

aha...rozumiem, ale tam chyba bedzie \(\frac{-1}{2}\)

[domino21]

dlaczego ten - ?

[malwik91]

\(\int \frac{x}{(x^2+1)^2} dx =\left(x^2+1 =t \\ 2x dx =dt \\ x dx =\frac{1}{2} dt \right)=\frac{1}{2} \int \frac{dt}{t^2} =\frac{1}{2} t^{-1} +C=\frac{1}{2t} +C=\frac{1}{2(x^2+1)}+C\)

wzór na całkę \(\int x^n dx=\frac{x^n+1}{n+1}+c\)
a tam było\(\frac{1}{2}\int t^{-2} dt=\frac{-1}{2}t^{-1}\)

[domino21]

uuu, masz rację

już poprawiam