Integral involving exponents

Question

How do we integrate $\int e^{C_1\frac{u^2+1}{u^2-1}} \ du\tag 1$ I could not find a proper substitution to convert it to a normal available form so that I can get a closed form of integration. $C_1$ is a constant

Answer 1

$\int e^{C_1\frac{u^2+1}{u^2-1}}~du$

$=\int e^{C_1\frac{u^2-1+2}{u^2-1}}~du$

$=e^{C_1}\int e^{\frac{2C_1}{u^2-1}}~du$

Let $u=\tanh x$ ,

Then $e^{C_1}\int e^{\frac{2C_1}{u^2-1}}~du$

$=e^{C_1}\int e^{\frac{2C_1}{\tanh^2x-1}}~d(\tanh x)$

$=e^{C_1}\int e^{-\frac{2C_1}{\text{sech}^2x}}\text{sech}^2x~dx$

$=e^{C_1}\int\dfrac{e^{-2C_1\cosh^2x}}{\cosh^2x}dx$

$=e^{C_1}\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^nC_1^n\cosh^{2n-2}x}{n!}dx$

$=e^{C_1}\int\left(\text{sech}^2x-2C_1+\sum\limits_{n=2}^\infty\dfrac{(-1)^n2^nC_1^n\cosh^{2n-2}x}{n!}\right)dx$

$=e^{C_1}\int\left(\text{sech}^2x-2C_1+\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{n+2}C_1^{n+2}\cosh^{2n+2}x}{(n+2)!}\right)dx$

For $n$ is any non-negative integer,

$\int\cosh^{2n+2}x~dx=\dfrac{(2n+2)!x}{4^{n+1}((n+1)!)^2}+\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sinh x\cosh^{2k+1}x}{4^{n-k+1}((n+1)!)^2(2k+1)!}+C$

This result can be done by successive integration by parts.

$\therefore e^{C_1}\int\left(\text{sech}^2x-2C_1+\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{n+2}C_1^{n+2}\cosh^{2n+2}x}{(n+2)!}\right)dx$

$=e^{C_1}\tanh x-2C_1e^{C_1}x+\sum\limits_{n=0}^\infty\dfrac{(-1)^nC_1^{n+2}e^{C_1}(2n+2)!x}{2^n((n+1)!)^3(n+2)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_1^{n+2}e^{C_1}(2n+2)!(k!)^2\sinh x\cosh^{2k+1}x}{2^{n-2k}((n+1)!)^3(n+2)(2k+1)!}+C$

$=e^{C_1}\tanh x+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_1^{n+2}e^{C_1}(2n+2)!(k!)^2\sinh x\cosh^{2k+1}x}{2^{n-2k}((n+1)!)^3(n+2)(2k+1)!}+\sum\limits_{n=-1}^\infty\dfrac{(-1)^nC_1^{n+2}e^{C_1}(2n+2)!x}{2^n((n+1)!)^3(n+2)}+C$

$=e^{C_1}\tanh x+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_1^{n+2}e^{C_1}(2n+2)!(k!)^2\sinh x\cosh^{2k+1}x}{2^{n-2k}((n+1)!)^3(n+2)(2k+1)!}-\sum\limits_{n=0}^\infty\dfrac{(-1)^nC_1^{n+1}e^{C_1}(2n)!x}{2^{n-1}(n!)^3(n+1)}+C$

$=e^{C_1}u+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_1^{n+2}e^{C_1}(2n+2)!(k!)^2u}{2^{n-2k}((n+1)!)^3(n+2)(2k+1)!(1-u^2)^{k+1}}-\sum\limits_{n=0}^\infty\dfrac{(-1)^nC_1^{n+1}e^{C_1}(2n)!\tanh^{-1}u}{2^{n-1}(n!)^3(n+1)}+C$

Or you can express in terms of Incomplete Bessel Functions:

$\int e^{C_1\frac{u^2+1}{u^2-1}}~du$

$=\int e^{C_1\frac{u^2-1+2}{u^2-1}}~du$

$=e^{C_1}\int e^{\frac{2C_1}{u^2-1}}~du$

Let $u=\tanh x$ ,

Then $e^{C_1}\int e^{\frac{2C_1}{u^2-1}}~du$

$=e^{C_1}\int e^{\frac{2C_1}{\tanh^2x-1}}~d(\tanh x)$

$=e^{C_1}\int e^{-\frac{2C_1}{\text{sech}^2x}}~d(\tanh x)$

$=e^{C_1}\int e^{-2C_1\cosh^2x}~d(\tanh x)$

$=e^{C_1(1-2\cosh^2x)}\tanh x-e^{C_1}\int\tanh x~d(e^{-2C_1\cosh^2x})$

$=e^{-C_1\cosh2x}\tanh x+4C_1\int e^{C_1(1-2\cosh^2x)}\sinh x\cosh x\tanh x~dx$

$=e^{-C_1\cosh2x}\tanh x+4C_1\int e^{-C_1\cosh2x}\sinh^2x~dx$

$=e^{-C_1\cosh2x}\tanh x+2C_1\int e^{-C_1\cosh2x}\cosh2x~dx-2C_1\int e^{-C_1\cosh2x}~dx$

$=e^{-C_1\cosh2x}\tanh x+C_1\int e^{-C_1\cosh2x}\cosh2x~d(2x)-C_1\int e^{-C_1\cosh2x}~d(2x)$

$=e^{-C_1\cosh2x}\tanh x+C_1J(C_1,1,2x)-C_1J(C_1,0,2x)+C$

$=ue^{C_1\frac{u^2+1}{u^2-1}}+C_1J(C_1,1,2\tanh^{-1}u)-C_1J(C_1,0,2\tanh^{-1}u)+C$

Answer 2

By replacing $u$ with $\sqrt{\frac{t-1}{t+1}}$ we get: $$\int \exp\left(C_1\frac{u^2+1}{u^2-1}\right)\,du=\int (t-1)^{-1/2}(t+1)^{-3/2}e^{-C_1t}\,dt$$ and the last integral over $[1,+\infty)$ can be evaluated in terms of Bessel K-functions.