Question

integrate (e^3alogex + e^3xlogea) dx​

Answer 1

Step-by-step explanation:

I hope this detailed explanation will help

integral of ( e^(3alog x)+ e^(3xlog a) ) dx

= integral of e^(3alog x)dx + integral of e^(3xlog a) dx

Let integral of e^(3alog x)dx = A

and integral of e^(3xlog a) dx = B

To find A

let 3alogx = p ....1

=> A = integral of e^(p) dp (dx/dp)

Let # refer to " integral of "

= # ( 1/3a) x e^(p) dp (since (dx/dp) = x )

= # (1/3a) e^( p/3a ) e^(p) dp (since x = e^( p/3a ) from ....1 )

= # (1/3a) e^[ (p + 3ap)/3a ] dp

= # e^[ (p + 3ap)/3a ] d(p + 3ap)/3a

A = e^[ (p + 3ap)/3a ] = e^[ (3alogx + 9a^2logx)/3a ]

on substituting p = 3alogx

To find B

B = # e^(3xlog a) dx

= # e^(3xlog a) dx

= # (1/3loga) e^(3xlog a) d(3xloga)

= (1/3loga) e^(3xlog a)

Thus the integral will be

e^[ (3alogx + 9a^2logx)/3a ] + (1/3loga) e^(3xlog a)

Answer 2

Hope you understand!!!