/
|
| dx
| ------------
|. x(x^2+4)
|
/
1
/
| dx
| ------------
| e^x+1
|
/
0
1
/
|
| ln(x+1) dx
/
0

1) $intlimits { frac{dx}{x( x^{2} +4)} } , dx$
$frac{1}{x( x^{2} +4)}= frac{A}{x} + frac{Cx+B}{ x^{2} +4}$
$frac{1}{x( x^{2} +4)}= frac{A( x^{2} +4)+x(Cx+B)}{x(x^{2} +4)}$
$1=A x^{2} +4A+C x^{2} +Bx$ ⇒
$left { {{4A=1, B=0} atop {A+C=0}} right.$
⇒ $A= frac{1}{4}, B=0, C= frac{-1}{4}$
⇒ $intlimits { frac{dx}{x( x^{2} +4)} } , dx= frac{1}{4} intlimits { frac{1}{x} } , dx - frac{1}{4} intlimits { frac{x}{ x^{2} +4} } , dx = frac{1}{4}lnx- frac{1}{8} intlimits { frac{1}{ x^{2} +4} } , d( x^{2} +4)=$ $frac{1}{4}lnx- frac{1}{8}ln( x^{2} +4)+C= frac{1}{8} ln( frac{ x^{2} }{ x^{2} +4})+C$

2) $intlimits^1_0 { frac{1}{ e^{x} +1} } , dx$ = $intlimits^1_0 { frac{1+e^{x}-e^{x}}{ e^{x} +1} } , dx$ = $intlimits^1_0 { frac{1+e^{x}}{ e^{x} +1} } , dx-intlimits^1_0 { frac{e^{x}}{ e^{x} +1} } , dx$ = $intlimits^1_0 {} , dx -intlimits^1_0 { frac{1}{ e^{x} +1} } , d(e^{x}+1)=x+ln(e^{x} +1) intlimits^1_0$ = $1+ln(e +1) -ln2=1+ln frac{e+1}{2}$

3) $intlimits^1_0 {ln(x+1)} , dx =ln(x+1)*x intlimits^1_0 - intlimits^1_0 { frac{x}{x+1} } , dx =$ $ln(1+1)-ln(1+0)- intlimits^1_0{ frac{x+1-1}{x+1} } , dx =$ $ln(2)-ln(1)- intlimits^1_0 { frac{x+1}{x+1} } , dx + intlimits^1_0 { frac{1}{x+1} } , dx =ln2-0+(-x+ln(x+1)) intlimits^1_0=$ $ln2-1+ln2=2ln2-1$