Let
ex=t⇒exdx=dt When,
x→−loge2 , then
t→21 When,
x→loge2 , then
t→2 ...........(i)
On applying integration by part method in Eq. (i), we get
amp;I=[tloge(t+1+t2)]1/22−∫1/22t+1+t2t(1+21+t22t)dtamp;=2loge(2+5)−21loge(21+5)−∫1/221+t2tdt =loge((21+5)1/2(2+5)2)−21∫1/221+t22tdt .............(ii)
Let
I1=∫1/221+t22tdt Let
1+t2=w 2tdt=dw When,
t→21 , then
w=45 When,
t→2 , then
w=5 I1amp;=∫5/45w1dwamp;=[2w]5/45amp;=2[5−25]=5 On substitute value of
I1 in Eq. (ii), we get
I=loge(1+52(2+5)2)−25