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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

Answers by Expert:

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.

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.)