Matematyka.pl

Mam problem z calkami :
1.\(\displaystyle{ \int{\frac{3x+2}{4x^2+4x+1}dx}}\)
2.\(\displaystyle{ \int{\frac{4x+7}{x^2+3x+3}dx}}\)
3.\(\displaystyle{ \int{\frac{x^4+x^3-2x^2-x+2}{x^2+2x}dx}}\)

Pierwsza "prawie" udalo mi sie zrobic, ale zawsze cos mi tam brakowalo do poprawnego wyniku...

\(\displaystyle{ 1.) t\frac{3x+2}{4x^2+4x+1}dx=\frac{3}{4}\int{4x+\frac{8}{3}}{4x^2+4x+1}dx=
\frac{3}{4}\int\frac{4x+2}{4x^2+4x+1}dx+\frac{3}{4}\int\frac{\frac{2}{3}}{4x^2+4x+1}dx=
\frac{3}{8}\int\frac{8x+4}{4x^2+4x+1}dx+\frac{1}{2}\int\frac{1}{(2x+1)^2}dx=}\)

podstawiamy \(\displaystyle{ t=2x+1, dt=2dx, dx=\frac{1}{2}dt}\)

\(\displaystyle{ =\frac{3}{8}\ln{|4x^2+4x+1|} + \frac{1}{4}\int\frac{1}{t^2}dt=\frac{3}{8}\ln{|4x^2+4x+1|} -\frac{1}{4t} + C = \frac{3}{8}\ln{|4x^2+4x+1|} -\frac{1}{4(2x+1)} + C}\)


\(\displaystyle{ 2.) t\frac{4x+7}{x^2+3x+3}dx=\int\frac{4x+6}{x^2+3x+3}dx+\int\frac{1}{x^2+3x+3}dx=
2\int\frac{2x+3}{x^2+3x+3}dx+\int\frac{1}{(x+\frac{3}{2})^2+\frac{3}{4}}dx=2\ln|x^2+3x+3|+
\frac{4}{3}\int\frac{1}{\frac{4}{3}(x+\frac{3}{2})^2+1}dx=
2\ln{|x^2+3x+3|}+\frac{4}{3}\int\frac{1}{(\frac{2x+3}{\sqrt{3}})^2+1}dx=}\)

podstawiamy \(\displaystyle{ t=\frac{2x+3}{\sqrt{3}}, dt=\frac{2}{\sqrt{3}}dx, dx=\frac{\sqrt{3}}{2}dt}\)

\(\displaystyle{ =2\ln{|x^2+3x+3|}+\frac{4}{3}\frac{\sqrt{3}}{2}\int\frac{1}{t^2+1}dt=
2\ln{|x^2+3x+3|}+\frac{2}{\sqrt{3}}\arctan{t}+C= 2\ln{|x^2+3x+3|}+\frac{2}{\sqrt{3}}\arctan{\frac{2x+3}{\sqrt{3}}+C}\)

\(\displaystyle{ 3.) t\frac{x^4+x^3-2x^2-x+2}{x^2+2x}dx=\int\frac{x^4+2x^3-x^3-2x^2-x+2}{x^2+2x}=
t\frac{x^4+2x^3}{x^2+2x}dx+\int\frac{-x^3-2x^2}{x^2+2x}dx+\int\frac{-x+2}{x^2+2x}dx=
t{x^2}dx-\int{x}dx+\int\frac{-x-1}{x^2+2x}dx+\int\frac{3}{x^2+2x}dx=\frac{1}{3}x^3-\frac{1}{2}x^2-
\frac{1}{2}\int\frac{2x+2}{x^2+2x}dx+\int\frac{3}{x(x+2)}dx=
\frac{1}{3}x^3-\frac{1}{2}x^2-
\frac{1}{2}\ln{|x^2+2x|}+\int\left(\frac{\frac{3}{2}}{x}-\frac{\frac{3}{2}}{x+2}\right)dx=
\frac{1}{3}x^3-\frac{1}{2}x^2-\frac{1}{2}\ln{|x^2+2x|}+
\frac{3}{2}\ln{|x|}-\frac{3}{2}\ln{|x+2|}+C=
\frac{1}{3}x^3-\frac{1}{2}x^2-\frac{1}{2}\ln{|x^2+2x|}+
\frac{3}{2}\ln{|\frac{x}{x+2}|}+C}\)

Jakby coś sie nie zgadzało to napisz bo możliwe że gdzieś sie pomyliłem