How can I integrate $\ln \left( x+\sqrt{1+x^2} \right) $?

Question

How can I integrate $$\ln \left( x+\sqrt{1+x^2} \right) $$ The problem is from IA Maron Calculus, Problem 4.3.17. The problem is under the heading 'Integration by parts'. I have tried it, but it just leads to 2 unsolvable integrals

Answer 1

Assuming the correct question is the edited one. With integration by parts: $$\int \ln\left(x+\sqrt{1+x^{2}}\right) \,\mbox{d}x = x\ln\left(x+\sqrt{1+x^{2}}\right)-\color{blue}{\int \frac{x}{\sqrt{1+x^2}}\,\mbox{d}x}$$ Since: $$\left(\ln\left(x+\sqrt{1+x^{2}}\right) \right)' =\left( \mbox{arcsinh}\,x \right)' = \frac{1}{\sqrt{1+x^2}}$$

For the blue integral, you can substitute $u=1+x^2$ and end up with:

$$\int \ln\left(x+\sqrt{1+x^{2}}\right) \,\mbox{d}x = x\ln\left(x+\sqrt{1+x^{2}}\right)-\color{blue}{\sqrt{1+x^2}}+C$$

Answer 2

As I wrote in the comment, the integral is not trivial. There has to be some error in the command.

But for the sake of completeness, here is the solution.

$$\log \left(x+\sqrt{x^4+1}\right) x-2 x+\frac{1}{2} \left(\sinh ^{-1}\left(x^2\right)-\frac{i \left(-3 i+\sqrt{3}\right) \tan ^{-1}\left(\frac{\left(3+i \sqrt{3}\right) x^8-2 i \sqrt{6-6 i \sqrt{3}} \sqrt{x^4+1} x^6+2 \sqrt{3} i \left(\sqrt{2-2 i \sqrt{3}} \sqrt{x^4+1}+2\right) x^4-2 i \sqrt{6-6 i \sqrt{3}} \sqrt{x^4+1} x^2+\sqrt{3} i-3}{\left(-7 i+\sqrt{3}\right) x^8+2 \left(i+\sqrt{3}\right) x^6+2 \left(-3 i+\sqrt{3}\right) x^4+2 \left(i+\sqrt{3}\right) x^2-i+\sqrt{3}}\right)}{\sqrt{6-6 i \sqrt{3}}}+\frac{\left(3-i \sqrt{3}\right) \tan ^{-1}\left(\frac{\left(3 i+\sqrt{3}\right) i x^8-2 i \sqrt{6+6 i \sqrt{3}} \sqrt{x^4+1} x^6+2 \sqrt{3} i \left(\sqrt{2+2 i \sqrt{3}} \sqrt{x^4+1}+2\right) x^4-2 i \sqrt{6+6 i \sqrt{3}} \sqrt{x^4+1} x^2+\sqrt{3} i+3}{\left(7 i+\sqrt{3}\right) x^8+2 \left(-i+\sqrt{3}\right) x^6+2 \left(3 i+\sqrt{3}\right) x^4+2 \left(-i+\sqrt{3}\right) x^2+i+\sqrt{3}}\right)}{\sqrt{6+6 i \sqrt{3}}}+\frac{\left(3 i+\sqrt{3}\right) \log \left(16 \left(x^4-x^2+1\right)^2\right)}{2 \sqrt{6+6 i \sqrt{3}}}+\frac{\left(-3 i+\sqrt{3}\right) \log \left(16 \left(x^4-x^2+1\right)^2\right)}{2 \sqrt{6-6 i \sqrt{3}}}-\frac{\left(3 i+\sqrt{3}\right) \log \left(\left(x^4-x^2+1\right) \left(\left(-3 i+\sqrt{3}\right) x^4-i \left(\sqrt{2+2 i \sqrt{3}} \sqrt{x^4+1}+2\right) x^2-3 i+\sqrt{3}-2 i \sqrt{2+2 i \sqrt{3}} \sqrt{x^4+1}\right)\right)}{2 \sqrt{6+6 i \sqrt{3}}}-\frac{\left(-3 i+\sqrt{3}\right) \log \left(\left(x^4-x^2+1\right) \left(\left(3 i+\sqrt{3}\right) x^4+i \left(\sqrt{2-2 i \sqrt{3}} \sqrt{x^4+1}+2\right) x^2+3 i+\sqrt{3}+2 \sqrt{2-2 i \sqrt{3}} i \sqrt{x^4+1}\right)\right)}{2 \sqrt{6-6 i \sqrt{3}}}\right)-\frac{\left(-3 i+\sqrt{3}\right) \tan ^{-1}\left(\frac{1}{2} \left(1-i \sqrt{3}\right) x\right)}{\sqrt{-6+6 i \sqrt{3}}}-\frac{\left(3 i+\sqrt{3}\right) \tan ^{-1}\left(\frac{1}{2} \left(1+i \sqrt{3}\right) x\right)}{\sqrt{-6-6 i \sqrt{3}}}$$

Answer 3

Notice the identity

$$\sqrt{\sinh^2(t)+1}=\cosh t$$ which hints the change of variable $x=\sinh t$ and $dx=\cosh t\,dt$.

The integral becomes

$$\int\log(\sinh t+\cosh t)\cosh t\,dt=\int t\cosh t\,dt.$$

Then by parts

$$t\,\sinh t-\int\sinh t\,dt=t\,\sinh t-\cosh t=x\text{ arsinh }x-\sqrt{x^2+1}.$$