CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

4ex+6ex9ex4exdx=Ax+Bloge(9e2x4)CA=_____,B=_____,C=_____

Open in App
Solution

Lets write
4ex+6ex=A(9eX4ex)+Bd(9ex4ex)dx
=A(9ex4ex)+B(9ex+4ex)
ee(9A+9B)+(4A+4B)ex
by comparing both sides, we get
9A+9B=4 of A+B=49B=49A
and 4A+4B=6
4A+4(49A)=6
4A+1694A=6
8A=6169
8A=34169
8A=389
A=388×9=1936
B=49(1936)
=16+1936=3536
1936(19x4ex)+3536(19x+4ex)(9ex4ex)
=1936dx+353619x+4ex19x4exdx
let 19x4ex=t
=(19x4ex)dx=dt
=I=1936x+3536log(192x4ex)+c
=1936x+3536log(192x4e2x)+c
=1936x+3536[log(192x4)loge2x]+c
=1936x+3536log(192x4)3536×2x+c
=8936x+3536log(192x4)+c

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Special Integrals - 1
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon