how can be calculated the definite integral of ln((|x|+1)/(x^2+1)) if limits of integration are -1,1? |x| is absolute value

Monday, January 13, 2014

how can be calculated the definite integral of ln((|x|+1)/(x^2+1)) if limits of integration are -1,1? |x| is absolute value

$\displaystyle{f{{\left({x}\right)}}}={\ln{}}$
(x+1)/(x^2+1)

$\displaystyle=\Rightarrow{f{{\left({x}\right)}}}={\ln{{\left({x}+{1}\right)}}}-$
ln(x^2+1)

$\displaystyle=\Rightarrow\int{f{{\left({x}\right)}}}=\int{\ln{{\left({x}+{1}\right)}}}-\int$
ln(x^2+1)

$\displaystyle{u}={x}+{1}=\Rightarrow{d}{u}=$
dx

$\displaystyle\int{\ln{{\left({x}+{1}\right)}}}{\left.{d}{x}\right.}=\int{\ln{{u}}}{d}{u}={u}\cdot{\ln{{u}}}-{u}+{c}$

$\displaystyle=\Rightarrow\int{\ln{{\left({x}+{1}\right)}}}{\left.{d}{x}\right.}={\left({x}+{1}\right)}\cdot{\ln{{\left({x}+{1}\right)}}}-{\left({x}+{1}\right)}$
......(1)

$\displaystyle\int{\ln{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle{u}={\ln{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}=\Rightarrow{d}{u}={2}\frac{{x}}{{{{x}}^{{2}}+{1}}}$
dx

$\displaystyle{d}{v}={\left.{d}{x}\right.}=\Rightarrow{v}=$
x

$\displaystyle\int{\ln{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}={u}\cdot{v}-\int$
v*du

$\displaystyle\int{\ln{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}={x}{\ln{{\left({{x}}^{{2}}+{1}\right)}}}-\int{2}\frac{{{x}}^{{2}}}{{{{x}}^{{2}}+{1}}}$
dx

$\displaystyle\int{\ln{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}={x}{\ln{{\left({{x}}^{{2}}+{1}\right)}}}-{2}\int\frac{{{x}}^{{2}}}{{{{x}}^{{2}}+{1}}}$
dx

$\displaystyle\int{\ln{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}={x}{\ln{{\left({{x}}^{{2}}+{1}\right)}}}-{2}{\left({x}-{\arctan{{\left({x}\right)}}}\right.}$

$\displaystyle\int{\ln{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}={x}{\ln{{\left({{x}}^{{2}}+{1}\right)}}}-{2}{x}-{2}{\arctan{{\left({x}\right)}}}+$
c.............(2)

$\displaystyle\int{\ln{{\left({x}+{1}\right)}}}-\int{\ln{{\left({{x}}^{{2}}+{1}\right)}}}=$
(x+1)ln(x+1) - (x+1) - xln(x^2+1) + 2x + 2arctan(x) +
C

$\displaystyle\int{\ln{{\left(\frac{{{x}+{1}}}{{{{x}}^{{2}}+{1}}}\right)}}}={\left({x}+{1}\right)}{\ln{{\left({x}+{1}\right)}}}-{x}{\ln{{\left({{x}}^{{2}}+{1}\right)}}}+$
2arctan(x) +x + 1 + C

$\displaystyle{\int_{{-{{1}}}}^{{1}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={\int_{{-{{1}}}}^{{1}}}$
ln((x+1)/(x^2+1)) dx = (2ln2-2ln2 +2arctan(1) + 1+1 - [ 0+ln(2) + 2arctan(-1) -1
+1}

$\displaystyle={2}{\arctan{{\left({1}\right)}}}+{2}-{\ln{{2}}}+$
2arctan(-1)

$\displaystyle={2}\cdot\frac{\pi}{{4}}+{2}-{\ln{{2}}}+$
2*-pi/4

$\displaystyle={2}-$
ln2

Help Notes

Monday, January 13, 2014