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# Calculus/Integral of rational function

Question
Dear Paul,

I have to calculate the primitive function of

integral(((1+sinh(x))/((1+cosh(x))dx)

I have tried approaching it this way:

Substituting
u=1+coshx
du=sinhx dx

integral(1+du)/u

but that is noway giving me a correct result.
So I figured the following

cosh²x - sinh²x=1
cosh²x=1+sin²x

sqrt of both sides gives me cosh²x=1+sinhx (not sure if i'm sinning on some math rules there).

This however also doesn't get me to the correct result...

The correct result is:
-x+ln(1+exp(x))² - 2/(exp(x)+1) +C

Hopefully you'll be able to help me out.
Thanks,
Bob

Questioner:Bob
Country:Limburg, Belgium
Category:Calculus
Private:No
Subject:Integral of rational function
Question:
Dear Paul,
I have to calculate the primitive function of
integral(((1+sinh(x))/((1+cosh(x))dx)
I have tried approaching it this way:

Substituting
u=1+coshx
du=sinhx dx
integral(1+du)/u
but that is noway giving me a correct result.

So I figured the following
cosh²x - sinh²x=1
cosh²x=1+sin²x
sqrt of both sides gives me cosh²x=1+sinhx (not sure if i'm sinning on some math rules there).

>>>>>>>>>>>> Yes, you are.

This however also doesn't get me to the correct result...
The correct result is:
-x+ln(1+exp(x))² - 2/(exp(x)+1) +C
Hopefully you'll be able to help me out.
Thanks,
Bob
…………………………………………………..
integral(((1+sinh(x))/((1+cosh(x))dx)

Try:
1 + sinh x  1 – cosh x
----------- ----------- =
1 + cosh x  1 – cosh x

1 + sinh x  – cosh x – sinh x cosh x
----------- ----------- =
1 – cosh² x

Now  cosh² – sinh² = 1, and

Cosh² – 1 = sinh²

That may help.

HOWEVER, your "look in the back of the book" approach may be even better:

1 + sinh x    1 + (exp x – exp(-x))/2
---------- =  -----------------------
1 + cosh x    1 + (exp x + exp(-x))/2

2 + exp x – exp(-x)
--------------------
2 + exp x + exp(-x)

NOW do a substitution:

2 + exp x – exp(-x)
-------------------- dx,
2 + exp x + exp(-x)

u = exp(x),  du = exp(x) dx,  dx = du/u

2 + u – 1/u   du
------------- ---
2 + u + 1/u    u

2 + u – 1/u
------------- du
2u + u² + 1

2u + u² – 1
--------------- du
u(2u + u² + 1)

2u + u² – 1
------------ du
u(u + 1)²

Now you have a routine partial-fractions exercise.  (Yes, I can hear you saying: Routine, he says!  HE doesn't have to do it.)

Set up:

u² + 2u – 1    A       B        C
----------- = ---- + ----- + --------
u(u + 1)²      u     u + 1   (u + 1)²

etc.

Calculus

Volunteer

#### Paul Klarreich

##### Expertise

All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

##### Experience

I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.

Education/Credentials
(See above.)