Odpowiedź :
Rozwiązanie:
[tex]1.[/tex]
[tex]$\int {x^{2} \arctan x} \, dx =\left|\begin{array}{ccc}f=\arctan x&dg=x^{2} \ dx\\df=\frac{1}{1+x^{2}} \ dx&g=\frac{1}{3}x^{3} \end{array}\right|=[/tex]
[tex]$\frac{1}{3}x^{3} \arctan x -\frac{1}{3} \int {\frac{x^{3}}{1+x^{2}} } \, dx =\frac{1}{3}x^{3} \arctan x -\frac{1}{3} \int {x-\frac{x}{x^{2}+1} } \, dx=[/tex]
[tex]$=\frac{1}{3} x^{3}\arctan x-\frac{1}{3}\int {x-\frac{1}{2} \cdot \frac{2x}{x^{2}+1} } \, dx =\frac{1}{3}x^{3} \arctan x-\frac{1}{6} x^{2}+\frac{1}{6} \ln |x^{2}+1| +C[/tex]
[tex]2.[/tex]
[tex]$\int {\frac{\ln x}{x^{2}} } \, dx =\left|\begin{array}{ccc}f=\ln x&dg=\frac{1}{x^{2}} \ dx\\df=\frac{1}{x} \ dx&g=-\frac{1}{x} \end{array}\right|=-\frac{\ln x}{x} +\int {\frac{1}{x^{2}} } \, dx =-\frac{\ln x}{x}-\frac{1}{x}+C[/tex]
[tex]3.[/tex]
[tex]$\int {\frac{2}{1+\sin x+\cos x} } \, dx[/tex]
Tutaj stosujemy podstawie Weierstrassa (uniwersalne) :
[tex]$t=\tan \frac{x}{2}[/tex]
[tex]$\frac{x}{2}=\arctan t[/tex]
[tex]$dx=\frac{2}{1+t^{2}} \ dt[/tex]
Ponadto:
[tex]$\sin x=\frac{2t}{1+t^{2}} \ , \ \cos x=\frac{1-t^{2}}{1+t^{2}}[/tex]
Mamy:
[tex]$\int {\frac{2}{1+\sin x+\cos x} } \, dx=\int {\frac{2}{1+\frac{2t}{1+t^{2}} +\frac{1-t^{2}}{1+t^{2}} } } \, \cdot \frac{2}{1+t^{2}} \ dt=[/tex]
[tex]$=4\int {\frac{1}{\frac{1+t^{2}-t^{2}+2t+1}{1+t^{2}} } } \cdot \frac{1}{1+t^{2}} \ dt=4\int {\frac{1+t^{2}}{2t+2} } \, \cdot \frac{1}{1+t^{2}} dt =2\int\ {\frac{1}{t+1} } \, dt=[/tex]
[tex]$=2\ln|t+1|+C=2\ln\Big|\tan\frac{x}{2} +1\Big|+C[/tex]
[tex]4.[/tex]
[tex]$\int {\frac{x^{3}}{1-x^{8}} } \, dx =\int {\frac{x^{3}}{1-(x^{4})^{2}} } \, dx =\left|\begin{array}{ccc}u=x^{4}\\du=4x^{3} \ dx\end{array}\right|=\frac{1}{4} \int {\frac{1}{1-u^{2}} } \, du=[/tex]
[tex]$=-\frac{1}{4} \int {\frac{1}{u^{2}-1} } \, du[/tex]
Tu korzystamy z dobrze znanego wzoru całkowego:
[tex]$\int {\frac{1}{x^{2}-a^{2}} } \, dx =\frac{1}{2a}\ln \Big|\frac{x-a}{x+a} \Big|+C[/tex]
Mamy:
[tex]$-\frac{1}{4} \int {\frac{1}{u^{2}-1} } \, du=-\frac{1}{8} \ln \Big|\frac{u-1}{u+1} \Big| +C=-\frac{1}{8} \ln \Big|\frac{x^{4}-1}{x^{4}+1} \Big|+C[/tex]
[tex]5.[/tex]
[tex]$\int {\frac{x^{2}-5x+9}{x^{2}+5x+6} } \, dx =\int {1+\frac{-10x+3}{x^{2}+5x+6} } \, dx =x+\int {\frac{-5(2x+5)+28}{x^{2}+5x+6} } \, dx =[/tex]
[tex]$=x-5\int {\frac{2x+5}{x^{2}+5x+6} } \, dx +28\int {\frac{1}{x^{2}+5x+6} } \, dx =[/tex]
[tex]$=x-5\ln|x^{2}+5x+6|+28\int {\frac{1}{\Big(x+\frac{5}{2} \Big)^{2}-\frac{1}{4} } } \, dx =\left|\begin{array}{ccc}u=x+\frac{5}{2} \\du= \ dx\end{array}\right|=[/tex]
[tex]$=x-5\ln |x^{2}+5x+6|+28\int {\frac{1}{u^{2}-\Big(\frac{1}{2} \Big)^{2}} } \, du=[/tex]
[tex]$=x-5\ln |x^{2}+5x+6|+28\ln \Big|\frac{u-\frac{1}{2} }{u+\frac{1}{2} } \Big|+C=x-5\ln |x^{2}+5x+6|+28\ln \Big|\frac{x+2 }{x+3 } \Big|+C[/tex]
[tex]6.[/tex]
[tex]$\int {\frac{8}{x(x+1)^{2}} } \, dx[/tex]
Ułamki proste:
[tex]$\frac{8}{x(x+1)^{2}} =\frac{A}{x} +\frac{B}{x+1} +\frac{C}{(x+1)^{2}}[/tex]
Mnożąc przez mianownik:
[tex]8=A(x+1)^{2}+Bx(x+1)+Cx[/tex]
[tex]8=A(x^{2}+2x+1)+B(x^{2}+x)+Cx[/tex]
[tex]8=x^{2}(A+B)+x(2A+B+C)+A[/tex]
Stąd:
[tex]A=8[/tex]
[tex]A+B=0 \iff B=-A=-8[/tex]
[tex]2A+B+C=0 \iff C=-2A-B=-16+8=-8[/tex]
Zatem całka wygląda tak:
[tex]$\int {\frac{8}{x(x+1)^{2}} } \, dx=\int {\frac{8}{x}-\frac{8}{x+1}-\frac{8}{(x+1)^{2}} } \, dx =[/tex]
[tex]$=8\ln |x|-8 \ln |x+1|+\frac{8}{x+1} +C[/tex]
[tex]7.[/tex]
[tex]$\int \frac{e^{-4x}}{4+e^{-4x}} \, dx =\left|\begin{array}{ccc}t=-4x\\dt=\ -4 \ dx\end{array}\right|=-\frac{1}{4}\int {\frac{e^{t}}{e^{t}+4} } \, dt =\left|\begin{array}{ccc}u=e^{t}+4\\du=e^{t} \ dt\end{array}\right|=[/tex]
[tex]$=-\frac{1}{4}\int {\frac{1}{u} } \, du=-\frac{1}{4}\ln|u|+C =-\frac{1}{4} \ln |e^{u}+4|+C=-\frac{1}{4}\ln|e^{-4x}+4|+C[/tex]