29Substitute 1 + xex = t
dt/dx = ex(x+1)
then the question is reduced to..
∫ dt / t2(t-1)..
which i think can be easily integrated using partial fractions.. : )
2 Answers
1let 1+ xe^x = t
x .e^x + e^x dx = dt
e^x ( x + 1 ) dx = dt
( x + 1 ) dx = dt / e^x
integral reduces to
∫dt t-1 ( t ) ^2
29