Matematyka.pl

Czy calka jest rozwiazana poprawnie?

\(\displaystyle{ I = \int_{}^{} ln(x^2+4)dx}\)
stosujac wzór: \(\displaystyle{ \int_{}^{} f(x) \cdot g'(x)dx = f(x) \cdot g(x) - \int_{}^{} f'(x) \cdot g(x)dx}\)

\(\displaystyle{ f(x)=ln(x^2+4)}\)
\(\displaystyle{ f'(x) = \frac{2x}{x^2+4}}\)
\(\displaystyle{ g'(x)=1}\)
\(\displaystyle{ g(x) = x}\)

\(\displaystyle{ I = ln(x^2+4) \cdot x - \int_{}^{} \frac{2x^2}{x^2+4} = ln(x^2+4) \cdot x - 2 \cdot \int_{}^{} \frac{x^2}{x^2(1+ \frac{4}{x^2} } =}\)
\(\displaystyle{ = ln(x^2+4) \cdot x - 2 \cdot \int_{}^{} \frac{1}{1+ \frac{4}{x^2} } = ln(x^2+4) \cdot x - 2arctg( \frac{2}{x})}\)

Nie do końca.

\(\displaystyle{ I = ln(x^2+4) \cdot x - 2\int_{}^{} \frac{x^2}{x^2+4}dx =}\)

\(\displaystyle{ 2\int_{}^{} \frac{x^2 +4-4}{x^2+4}dx=2 \int dx -2\int \frac{1}{\left( \frac{x}{2} \right) ^{2} +1}dx=2x-4\ arctg\left( \frac{x}{2} \right)}\)

\(\displaystyle{ I = ln(x^2+4) \cdot x -2x+4\ arctg\left( \frac{x}{2} \right)}\)

ok, dzieki wielkie za pomoc