How to find $\int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$

Question

$$I=\int x.\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$$

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Try 1: Put $z= \ln(x+\sqrt{1+x^2})$, $\mathrm dz=1/\sqrt{1+x^2}\mathrm dx$ $$I=\int \underbrace{x}_{\mathbb u}\underbrace{z}_{\mathbb v}\mathrm dz=x\int zdz-\int (z^2/2)\mathrm dz\tag{Wrong}$$

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Try 2: Put $z= x+\sqrt{1+x^2}$ $$\implies x-z =\sqrt{1+x^2}\implies x^2+z^2-2xz =1+x^2\implies x =\frac{z^2-1}{2z}$$ $$\mathrm dz =\left(1+\frac{x}{\sqrt{1+x^2}}\right)\mathrm dx =\frac{z\mathrm dx}{x-z}=\frac{-2z^2\mathrm dx}{1+z^2}$$ $$I =\int\frac{(z^2-1)\ln z}{2z}.\frac{(1+z^2)\mathrm dz}{-2z^2}$$ $$=\int\frac{(z^4-1)\mathrm dz}{4z^3} =\frac14\int\left(z-\frac1{z^3}\right)\mathrm dz =z^2/2+2/z^2+C\tag{Wrong}$$

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Try 3: Put $z =\sqrt{1+x^2},\mathrm dx =x/\sqrt{1+x^2}\mathrm dx$ $$I =\int \ln(x+z)\mathrm dz =\int \ln(z+\sqrt{z^2-1})\mathrm dz$$ Don't know how to solve this integral.

[Note that if I take $u=z+\sqrt{z^2-1}$, it is $=\sqrt{1+x^2}+\sqrt{1+x^2-1}=x+\sqrt{1+x^2}$; same as first try.]

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What's wrong in try 1 & 2, how to further solve try 3 and the best method to solve this question?

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Update: Sorry, I don't know hyperbolic/inverse hyperbolic trigonometry.

Answer 1

Continuation of Try 1:

$z=\ln(x+\sqrt{1+x^2})\implies e^{z}=x+\sqrt{1+x^2}\implies(e^z-x)^2=1+x^2\implies$

$\;\;\;e^{2z}-2xe^z=1\implies2xe^z=e^{2z}-1\implies x=\frac{1}{2}(e^z-e^{-z}),$

so $I=\int\frac{1}{2}z(e^z-e^{-z})\;dz=\frac{1}{2}[\int ze^z \;dz-\int ze^{-z}\;dz$]. $\;\;$ Now use integration by parts.

[Notice that we could have used $\frac{1}{2}(e^z-e^{-z})=\sinh z$].

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Continuation of Try 2:

$z=x+\sqrt{1+x^2}\implies z-x=\sqrt{1+x^2}\implies z^2-2xz+x^2=1+x^2\implies 2xz=z^2-1\implies$ $x=\frac{1}{2}(z-\frac{1}{z})\implies dx=\frac{1}{2}(1+\frac{1}{z^2})dz$.

Then $\displaystyle I=\int\frac{\frac{1}{2}(z-\frac{1}{z})\ln z}{z-\frac{1}{2}(z-\frac{1}{z})}\frac{1}{2}\left(1+\frac{1}{z^2}\right)dz=\frac{1}{2}\int\frac{(z^2-1)\ln z}{z^2+1}\cdot\frac{z^2+1}{z^2}\;dz$

$\;\;\;\displaystyle=\frac{1}{2}\int\left(1-\frac{1}{z^2}\right)\ln z\;dz.$ $\;\;$Now use integration by parts.

Answer 2

$$ \int \ln(z+\sqrt{z^2-1})dz = \int \mathrm{arcosh}(z) dz\,\,\,z>1. $$

the equality holds because for real x, $z = \sqrt{x^2+1}>1$. then using the fact $$ \int \mathrm{arcosh}(z) dz = z\mathrm{arcosh}(z) +\sqrt{z^2-1} + C $$

then sub back in your substitution.

$\textbf{brief derivation of the arcosh relation}$

$$ y = \cosh(x) \implies \mathrm{arcosh}(y) = x $$

now $$ \cosh(x) = \frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2} = y \tag{*} $$

I can re-write the expression in Eq.(*) with $\mathrm{e}^{x} = t$ as

$$ \cosh(x) = \frac{t+1/t}{2} = \frac{1}{t}\frac{t^2+1}{2} = y $$ so solving for $t$ we find

$$ t^2-2yt+1 = 0 \implies t = \frac{2y\pm\sqrt{4y^2-4}}{2} = y\pm\sqrt{y^2-1} $$

therefore $$ \mathrm{e}^{x} = t \implies x = \ln\left(y\pm\sqrt{y^2-1}\right) = \mathrm{arcosh}(y) $$ since we are looking for real valued function then we can ignore the minus sign.

At least in my head anyway.

Answer 3

Hint :

Use IBP by setting $u=x$ and $dv=\dfrac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\ dx$. Note that: $\text{arsinh}\ x=\ln(x+\sqrt{1+x^2})$ and $\dfrac{d}{dx}\text{arsinh}\ x=\dfrac{1}{\sqrt{1+x^2}}$, then \begin{align} \int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\ dx&=x\left[\int\text{arsinh}\ x\ d(\text{arsinh}\ x)\right]-\int\left[\int\text{arsinh}\ x\ d(\text{arsinh}\ x)\right]\ dx. \end{align} The rest part, you may refer to Wikipedia: hyperbolic function.

Answer 4

Hint: $$ \int x\frac{\ln({x+\sqrt{1+x^2})}}{\sqrt{1+x^2}}dx=\int\ln({x+\sqrt{1+x^2})}d\sqrt{1+x^2} $$ and $$ (\ln({x+\sqrt{1+x^2})})'=\frac{1}{\sqrt{1+x^2}}. $$