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Đặt f(x)=(x+1−1x)ex+1x
t=x+1x⇔x2−tx+112≤x≤2⇔x=[t−√t2−42≤1t+√t2−42≥1x=12⇒t=52x=1⇒t=2x=2⇒t=52
Suy ra: dx=[12(1−t√t2−4)dt,12≤x≤112(1+t√t2−4)dt1≤x≤2
Từ đó: I=2∫12f(x)dx=1∫12f(x)dx+2∫1f(x)dx
I=122∫52(t−√t2−42+1−t+√t2−42)et.(√t2−4−t√t2−4)dt−−122∫52(t+√t2−42+1−t−√t2−42)et.(√t2−4+t√t2−4)dt=−1252∫2(1−√t2−4)(1−t√t2−4)etdt++1252∫2(1+√t2−4)(1+t√t2−4)etdt
I=52∫2(√t2−4+t√t2−4)etdt=52∫2√t2−4etdt+52∫2tet√t2−4dt
Đặt u=et⇒du=etdt
dv=t√t2−4⇒v=√t2−4
52∫22t√t2−4etdt=et√t2−4]522−52∫2√t2−4etdt
1)I=ln3∫0dx√ex+12)J=2∫0xe−x2dx
⇒52∫2√t2−4etdt+52∫2t√t2−4etdt=32e52
ĐS : I=32e52
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