Question

$\int \frac{\left(ln\right(x)-1{)}^2}{\left(l{n}^2\right(x)+1{)}^2}dx$

Answer 1

Expand:
$=\int (\frac{l{n}^2\left(x\right)}{l{n}^4\left(x\right)+2l{n}^2\left(x\right)+1}-\frac{2ln\left(x\right)}{l{n}^4\left(x\right)+2l{n}^2\left(x\right)+1}+\frac{1}{l{n}^4\left(x\right)+2l{n}^2\left(x\right)+1})dx$
Apply linearity:
$=\int \frac{l{n}^2\left(x\right)}{l{n}^4\left(x\right)+2l{n}^2\left(x\right)+1}dx-2\int \frac{ln\left(x\right)}{l{n}^4\left(x\right)+2l{n}^2\left(x\right)+1}dx+\int \frac{1}{l{n}^4\left(x\right)+2l{n}^2\left(x\right)+1}dx$
Now solving:
$\int \frac{l{n}^2\left(x\right)}{l{n}^4\left(x\right)+2l{n}^2\left(x\right)+1}dxSubstituteu=ln\left(x\right)\to \frac{du}{dx}=\frac{1}{x}usex=e^u,l{n}^2\left(x\right)=u^2,l{n}^4\left(x\right)=u^4=\int \frac{u^2{e}^u}{u^4+2u^2+1}du$
Factor the denominator:
$=\int \frac{u^2{e}^u}{(u-i)(u+i{)}^2}du=\int \frac{u^2{e}^u}{(u-i{)}^2(u+i{)}^2}du$
Perform partial fraction decomposition:
$=\int (\frac{i{e}^u}{4(u+i)}+\frac{e^u}{4(u+i{)}^2}-\frac{i{e}^u}{4(u-i)}+\frac{e^u}{4(u-i{)}^2})du$
Apply linearity:
$=\frac{i}{4}\int \frac{e^u}{u+i}du-\frac{1}{4}\int \frac{e^u}{(u+i{)}^2}du-\frac{i}{4}\int \frac{e^u}{u-i}du-\frac{1}{4}\int \frac{e^u}{(u-i{)}^2}du$
Now solving:
$\int \frac{e^u}{u+i}du$
Use previous result:
$=e^{-i}{E}_i(u+i)$
Now solving:
$\int \frac{e^u}{(u+i{)}^2}du$
Use previous result:
$=e^{-i}{E}_i(u+i)-\frac{e^u}{u+i}\frac{i}{4}\int \frac{e^u}{u+i}du-\frac{1}{4}\int \frac{e^u}{(u+i{)}^2}du-\frac{i}{4}\int \frac{e^u}{u-i}du-\frac{1}{4}\int \frac{e^u}{(u-i{)}^2}du=\frac{i{e}^{-i}{E}_i(u+i)}{4}-\frac{e^{-i}{E}_i(u+i)}{4}-\frac{e^{i}i{E}_i(u-i)}{4}-\frac{e^i{E}_i(u-i)}{4}+\frac{e^u}{4(u+i)}+\frac{e^u}{4(u-i)}$
Undo substitution $u=ln\left(x\right),use{e}^{ln\left(x\right)}=x$:
$\frac{i{e}^{-i}{E}_i\left(ln\right(x)+i)}{4}-\frac{e^{-i}{E}_i\left(ln\right(x)+i)}{4}-\frac{e^{i}i{E}_i\left(ln\right(x)-i)}{4}-\frac{e^i{E}_i\left(ln\right(x)-i)}{4}+\frac{x}{4\left(ln\right(x)+i)}$
Plug in solved integrals:
$\int \frac{l{n}^2\left(x\right)}{l{n}^4\left(x\right)+2l{n}^2\left(x\right)+1}dx-2\int \frac{ln\left(x\right)}{l{n}^4\left(x\right)+2l{n}^2\left(x\right)+1}dx+\int \frac{1}{l{n}^4\left(x\right)+2l{n}^2\left(x\right)+1}dx=\frac{ix}{2\left(ln\right(x)+i)}-\frac{ix}{2\left(ln\right(x)-i)}$
The problem is solved:
$\int \frac{\left(ln\right(x)-1{)}^2}{\left(l{n}^2\right(x)+1{)}^2}dx=\frac{ix}{2\left(ln\right(x)+i)}-\frac{ix}{2\left(ln\right(x)-i)}+C$
Rewrite/simplify:
$=\frac{x}{l{n}^2\left(x\right)+1}+C$