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

1)[latex] \int\limits { \frac{dx}{x( x^{2} +4)} } \, dx [/latex] [latex] \frac{1}{x( x^{2} +4)}= \frac{A}{x} + \frac{Cx+B}{ x^{2} +4} [/latex] [latex] \frac{1}{x( x^{2} +4)}= \frac{A( x^{2} +4)+x(Cx+B)}{x(x^{2} +4)} [/latex] [latex]1=A x^{2} +4A+C x^{2} +Bx[/latex]  ⇒  [latex] \left \{ {{4A=1, B=0} \atop {A+C=0}} \right. [/latex] ⇒[latex]A= \frac{1}{4}, B=0, C= \frac{-1}{4} [/latex] ⇒ [latex] \int\limits { \frac{dx}{x( x^{2} +4)} } \, dx= \frac{1}{4} \int\limits { \frac{1}{x} } \, dx - \frac{1}{4} \int\limits { \frac{x}{ x^{2} +4} } \, dx = \frac{1}{4}lnx- \frac{1}{8} \int\limits { \frac{1}{ x^{2} +4} } \, d( x^{2} +4)= [/latex][latex] \frac{1}{4}lnx- \frac{1}{8}ln( x^{2} +4)+C= \frac{1}{8} ln( \frac{ x^{2} }{ x^{2} +4})+C [/latex] 2) [latex]\int\limits^1_0 { \frac{1}{ e^{x} +1} } \, dx [/latex]=[latex]\int\limits^1_0 { \frac{1+e^{x}-e^{x}}{ e^{x} +1} } \, dx[/latex] =[latex]\int\limits^1_0 { \frac{1+e^{x}}{ e^{x} +1} } \, dx-\int\limits^1_0 { \frac{e^{x}}{ e^{x} +1} } \, dx [/latex]= [latex]\int\limits^1_0 {} \, dx -\int\limits^1_0 { \frac{1}{ e^{x} +1} } \, d(e^{x}+1)=x+ln(e^{x} +1) \int\limits^1_0[/latex]=[latex]1+ln(e +1) -ln2=1+ln \frac{e+1}{2}[/latex] 3)[latex] \int\limits^1_0 {ln(x+1)} \, dx =ln(x+1)*x \int\limits^1_0 - \int\limits^1_0 { \frac{x}{x+1} } \, dx =[/latex][latex]ln(1+1)-ln(1+0)- \int\limits^1_0{ \frac{x+1-1}{x+1} } \, dx =[/latex][latex]ln(2)-ln(1)- \int\limits^1_0 { \frac{x+1}{x+1} } \, dx + \int\limits^1_0 { \frac{1}{x+1} } \, dx =ln2-0+(-x+ln(x+1)) \int\limits^1_0=[/latex][latex]ln2-1+ln2=2ln2-1[/latex]