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Evaluate $$\int _0 ^1 \frac{e^{ax}-e^{bx}}{(e^{ax}+1)(e^{bx}+1)} \mathrm{d}x$$ where $a$ and $b$ are constant. I attempt to factor the fraction but then have no ideas where to go. Please help me. Thanks!

S L
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user1500178
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1 Answers1

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$$\int _0 ^1 \frac{e^{ax}-e^{bx}}{(e^{ax}+1)(e^{bx}+1)} \mathrm{d}x$$

= $$\int _0 ^1 \frac{(e^{ax}+1) -(e^{bx}+1)}{(e^{ax}+1)(e^{bx}+1)} \mathrm{d}x$$

= $$\int _0 ^1 \frac{1}{(e^{bx}+1)} \mathrm{d}x$$ + $$\int _0 ^1 \frac{1}{(e^{ax}+1)} \mathrm{d}x$$

now for first integral substitute $(e^{bx}+1)$=u and for second integral substitute $(e^{ax}+1)$=v

now

= $$\int _2 ^{1+e^b}\frac{1}{b(u-1))}\mathrm{d}u$$ +$$\int _2 ^{1+e^a}\frac{1}{a(v-1))}\mathrm{d}v$$

i will leave it there hope you can manage rest if not ask i will type the rest

Mr. Math
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    I would have liked to see whether OP could have come up with this, using the hint in the comments. It would have been a learning experience. – Gerry Myerson Sep 22 '13 at 06:22
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    Instead we have an example of the "shopping at MSE experience". (@Jam: the upper bounds in the last integrals are incorrect.) – Did Sep 22 '13 at 08:01
  • @Did thanks for the man i should have paid more attention and i have edited thanks – Mr. Math Sep 22 '13 at 15:26
  • @GerryMyerson first of all i always don't read comments because there is too much bickering so i don't pay attention second if you had a left an answer saying as a hint and write what you have written in comments i wouldn't have bothered writing an answer. I for one feel there are too many opinions on how to write answer and when to write answer on MSE by a handful of users. I have no time for time paying attention to those as my time is better utilised on trying to help another student. – Mr. Math Sep 22 '13 at 15:30
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    Actually, I did figure out the partial fractions at first. The substitution is a little trickier. Do the substitution as @Jan suggested and then apply partial fraction again, I got the result $log \frac{e+1}{2}(\frac{1}{a}-\frac{1}{b})$. Many thanks to both you guys. I really enjoy the learning experience here. – user1500178 Sep 22 '13 at 15:41
  • We're all trying to help students here, Jam, it's just a question of whether it's more helpful in the long run to give 90% of the answer, or to give 10% and encourage the student to work out the rest. – Gerry Myerson Sep 22 '13 at 23:39