Solving a tricky integral with e.

A few notes:
1. You can remove the absolute value from the final answer since it is always positive.
2. You can factor out an e^(-4x) in the log and when you do this you get a somewhat "nicer" answer:
-4x + 11/15 ln (6e^(5x) )
3. You could also do this problem by multiplying numerator and denominator first by e^(4x) and then doing basically the same steps. Kind of the same thing but some may prefer this method.

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[owlsmath]

✋A few notes:
1. You can remove the absolute value from the final answer since it is always positive.
2. You can factor out an e^(-4x) in the log and when you do this you get a somewhat "nicer" answer:
-4x + 11/15 ln (6e^(5x) )
3. You could also do this problem by multiplying numerator and denominator first by e^(4x) and then doing basically the same steps. Kind of the same thing but some may prefer this method.

[mathsplus]

At the beginning I thought u substitution wouldn't be possible, this method is very effective with the a, b coefficients and a sum. Brilliant how it simplifies fast by the input of algebra.
Fantastic problem and solution ❤👍🙌

[prollysine]

Dear theacher ! Bruteforce solution : ln|e^(-4x)+6e^x| - (4/15)*ln|1+6e^(5x)| + C /x -->0, y=1, 42 *** x-->1, y=0, 97/

[theelk]

I solved this by finding an “integrating factor” e^t and multiplying the top and the bottom and taking the derivative of the bottom so I had (t-4)/-4=6(t+1)/-2 and solved for t=-16/11 so that with substitution I’d end up with the integral of 11/15u