Tính tích phân: $I=\int_e^{e^2} \frac{1-lnx}{lnx(x+lnx)}dx$

Tính tích phân: I=\int_e^{e^2}\frac{1-lnx}{lnx(x+lnx)}dx.

Lời giải.

Ta có
I=\int_e^{e^2}\frac{1-lnx}{lnx(x+lnx)}dx=\int_e^{e^2} \frac{x-x.lnx}{xlnx(x+lnx)}dx
=\int_e^{e^2}\frac{x+lnx-(x+1).lnx}{x.lnx.(x+lnx)}.dx=\int_e^{e^2}\frac{d(lnx)}{lnx}-\int_e^{e^2}\frac{1+1/x}{x+lnx}dx
=ln(lnx)|_e^{e^2}-\int_e^{e^2}\frac{d(x+lnx)}{x+lnx}dx
=ln2-ln(x+lnx)|_e^{e^2}
=ln2-ln(e^2+2)+ln(e+1).
Đáp số I=ln(\frac{2e+2}{e^2+2}).

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s