wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The value of the definite integral 11dx(1+ex)(1+x2) is

A
π2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
π4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
π8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
π16
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B π4
I=11dx(1+ex)(1+x2)(i)
Using the property: baf(x) dx=baf(a+bx) dx
=11dx1+ex.11+x2
I=11exdx(1+ex)(1+x2)(ii)
Adding (i) and (ii), we get
2I=11(1+ex)dx(1+ex)(1+x2)
=11dx(1+x2)
We know that,
aaf(x) dx=⎪ ⎪ ⎪⎪ ⎪ ⎪2a0f(x) dx, if f(x) is even0, if f(x) is odd
=210dx(1+x2)
I=10dx(1+x2)=tan1(1)=π4

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon