nauka obliczania całek

[studencina]

moglby ktos napisac rozwiazania tych całek, nie chodzi mi o wyniki czy podpowiedzi tylko o całe rozwiazania bo musze załapać jak to sie robi, z góry ogromne podziekowania

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

2.\(\displaystyle{ \int\sin^2xcosxdx}\)

3.\(\displaystyle{ \int\frac{3x}{sqrt (x+1)}dx}\)

4.\(\displaystyle{ \int\frac{x^3}{x^2-1}dx}\)

5.\(\displaystyle{ \int\frac{x}{1+x^2}dx}\)

6. \(\displaystyle{ \int_{0}^{1}\frac{1}{sqrt{x}}dx}\)

7. \(\displaystyle{ \int_{0}^{1}\ x^2cosxdx}\)

8. \(\displaystyle{ \int_{0}^{1}\ xe^xdx}\)

9. \(\displaystyle{ \int_{1}^{e}\ xlnxdx}\)

[el payaco]

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

PODSTAWIENIE:
\(\displaystyle{ t=1+4x^{2}\\ dt=8x dx\\ dx=\frac{dt}{8x}dt}\)

Narazie tyle bo na więcej nie mam czasu

[sushi]

2)

sinx==t
cosx dx= dt
i \(\displaystyle{ \int{ \sin^2 x \cos x dx}= \int {t^2 dt}==\frac{t^3}{3}+C==\frac{\sin^3 x}{3}+C}\)

[ Dodano: 22 Wrzesień 2006, 12:17 ]
3)
\(\displaystyle{ \sqrt{x+1}=t}\)
\(\displaystyle{ x+1=t^2}\), \(\displaystyle{ x=t^2-1}\)
dx= 2t dt

\(\displaystyle{ \int{\frac{3x}{\sqrt{x+1}}dx}= \int{ \frac{3(t^2-1)}{t}(2t dt)} = \int{(6t^2-6)}dt= 2t^3-6t +C= 2 (\sqrt{x+1})^3- 6 \sqrt{x+1}+C}\)

[ Dodano: 22 Wrzesień 2006, 12:24 ]
4)

\(\displaystyle{ x^2-1==t}\),\(\displaystyle{ x^2=t+1}\)

\(\displaystyle{ 2x dx=dt}\)
\(\displaystyle{ xdx=\frac{dt}{2}}\)

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

[ Dodano: 22 Wrzesień 2006, 12:38 ]
5)

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

\(\displaystyle{ \int{ \frac{x dx}{1+x^2}}= \int{\frac{1}{t} \frac{dt}{2}}= \int{{\frac{dt}{2t}}}= \frac{\ln t}{2}+C= \frac{ \ln (1+x^2)}{2}+C}\)

[ Dodano: 22 Wrzesień 2006, 12:44 ]
6)
\(\displaystyle{ \int{\frac{dx}{\sqrt{x}}}= 2 \sqrt{x}+C}\)
\(\displaystyle{ \int^{1}_{0}{\frac{dx}{\sqrt{x}}}=2 \sqrt{x}|^{1}_{0}= 2 \sqrt{1}- 2 \sqrt{0} =2}\)

[ Dodano: 22 Wrzesień 2006, 13:14 ]
7)
całkowanie przez części

\(\displaystyle{ \int {u \cdot v'}= u \cdot v - \int{u' \cdot v}}\)
u=x*x , u'= 2x
v'= cos x , v= sinx

\(\displaystyle{ \int_{0}^{1} {x^2 \cdot \cos x}dx= x^2 \cdot \sin x |_{0}^{1} - \int_{0}^{1} {2x \cdot \sin x}dx = [1^2 \cdot \sin 1- 0^2 \cdot \sin 0]- 2 \int{x \cdot \sin x}dx====}\)

u=x , u'= 1
v'= sin x , v= - cosx
\(\displaystyle{ \int{x \cdot \sin x}dx= - x \cdot \cos x + \int{ \cos x}= - x \cdot \cos x + \sin x +C}\)

\(\displaystyle{ ==== \sin 1- 2[- x \cdot \cos x + \sin x]|_{0}^{1} = \sin 1 + [ 2x \cdot \cos x -2 \sin x]|_{0}^{1}= \\ = \sin 1 +[(2 \cos 1 - 2 \sin 1)- (0 \cos 0 - 2 \sin 0)] = 2 \cos 1 - \sin 1}\)

[ Dodano: 22 Wrzesień 2006, 13:19 ]
8)

przez częsci u=x ;v'= e^{x}
u'=1 v= e^{x}

wyjdzie \(\displaystyle{ (x \cdot e^x- e^x)|_{0}^{1}}\)

[el payaco]

9.\(\displaystyle{ \int_{1}^{e}x\ln{x}dx=\langle u=\ln{x}\ \ u^{'}=\frac{1}{x}; v^{'}=x\ \ v=\frac{1}{2}x^{2} \rangle = (\frac{1}{2}x^{2}\ln{x})|_{1}^{e}-\int_{1}^{e}\frac{1}{x} \frac{1}{2}x^{2}=\frac{e^{2}}{2}-\frac{1}{4}e^{2}-(\frac{1}{4}x^{2})|_{1}^{e}=\frac{e^{2}}{2}-\frac{e^{2}}{4}+\frac{1}{4}=\frac{1}{4}(e^{2}+1)}\)

[sushi]

9)
u= ln x, v'=x
\(\displaystyle{ u'= \frac{1}{x}}\), \(\displaystyle{ v= \frac{x^2}{2}}\)

po obliczeniu całki wyjdzie
\(\displaystyle{ [\frac{ x^2 \cdot \ln x}{2} - \frac{x^2}{4} ] |_{1}^{e}}\)