J johnq2k7 New member Joined Feb 10, 2009 Messages 46 Mar 1, 2009 #1 integral of tanh(ln(x) dx: work shown: from table of integration of hyberbolic forms: integral of tanh(u) du= ln cosh(u) +C therefore i got (1/x)* ln (cosh(ln(x)) +C as my answer why is this wrong? please help
integral of tanh(ln(x) dx: work shown: from table of integration of hyberbolic forms: integral of tanh(u) du= ln cosh(u) +C therefore i got (1/x)* ln (cosh(ln(x)) +C as my answer why is this wrong? please help
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Mar 1, 2009 #2 \(\displaystyle \int tanh(ln(x))dx\) Rewrite as \(\displaystyle \int \frac{x^{2}-1}{x^{2}+1}dx\) \(\displaystyle \int\left(1-\frac{2}{1+x^{2}}\right)dx\) \(\displaystyle =x-2\int\frac{1}{1+x^{2}}dx\) \(\displaystyle x-2tan^{-1}(x)\)
\(\displaystyle \int tanh(ln(x))dx\) Rewrite as \(\displaystyle \int \frac{x^{2}-1}{x^{2}+1}dx\) \(\displaystyle \int\left(1-\frac{2}{1+x^{2}}\right)dx\) \(\displaystyle =x-2\int\frac{1}{1+x^{2}}dx\) \(\displaystyle x-2tan^{-1}(x)\)
