Integrate (logx)/(x(1+logx)(2+logx)) ?

Question

Integrate (logx)/(x(1+logx)(2+logx)) ?

1 Answer

Impact of this question

4823 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-20T13:23:22

$intlogx/(x(1+logx)(2+logx))dx=-log|logx+1|+2log|logx+2|+C$

Explanation:

$I=intlogx/(x(1+logx)(2+logx))dx$
Let $u=logx$ so $du=1/xdx$
$I=intu/((u+1)(u+2))du$
Let's examine $u/((u+1)(u+2))$ and its partial fraction decomposition:
$u/((u+1)(u+2))=A/(u+1)+B/(u+2)$
$u=A(u+2)+B(u+1)$
$u=-2->B=2$
$u=-1->A=-1$
Therefore, $u/((u+1)(u+2))=-1/(u+1)+2/(u+2)$
$I=int(-1/(u+1)+2/(u+2))du$
$I=-int1/(u+1)du+2int1/(u+2)du$
$I=-log|u+1|+2log|u+2|+C$
$I=-log|logx+1|+2log|logx+2|+C$

Science

Math

Humanities

... and beyond

Integrate (logx)/(x(1+logx)(2+logx)) ?

1 Answer

Explanation:

Related questions

Impact of this question