# Intergrate $\text{arccosh}(\sqrt{x^2 + 1})$ via hyperbolic substituation.

Sorry for the bad formatting, I have never done this before. I have an exam coming up in a few days and questions of the integration via hyperbolic substitution are the only ones that seem to stump me. How I attempted this question was:

let: $$x = \sinh(\theta)$$ therefore $$\theta = \text{arcsinh} (x), dx = d\theta\cosh(\theta)$$

thus: $$\int \text{arccosh}(\sqrt{x^2+1})dx = \int \text{arccosh}\sqrt{\sinh^2\theta+1}dx = \int(\text{arccosh}\left(\sqrt{\cosh^2\theta}\right)\cosh\theta d\theta$$ Which simplifies down to: $$\int\theta \cosh\theta d\theta$$. Using integration by parts, I got:

$$\int\theta \cosh(\theta)d\theta = \theta \sinh(\theta)-\sinh(\theta)+c$$

Subbing in $$\theta = \text{arcsinh}(x)$$, I finally get an answer of: $$x\text{arcsinh}(x) - x+c$$ However the answer is $$x\text{arcsinh}(x) - \sqrt{x^2+1} + c$$. I know I have made a mistake with my working out, however, I feel like I have a lack of understanding regarding hyperbolic substitutions so I cant understand where the mistake is located. Is it when I subbed in $$\text{arcsinh}(x) = \theta$$ or somewhere earlier?

Thanks !

• The IBP is wrong, but there is another problem: $\sqrt{\cosh^2 \theta}=\cosh(\theta)$, because $\cosh$ is always nonnegative. But then $\mathrm{arccosh}(\cosh(\theta))=|\theta|$, and not simply $\theta$. Commented Jul 15, 2023 at 13:29

I think you made a mistake during the integration by parts: $$\int\theta \cosh \theta d\theta = \theta \sinh \theta-\int\sinh \theta d\theta=\theta \sinh \theta-\cosh \theta+k$$ Going back to the $$x$$ world, you have: $$\theta \sinh \theta=x \text{arcsinh }x$$ $$\cosh \theta=\sqrt{1+\sinh^2\theta}=\sqrt{1+x^2}$$ Hence the result is: $$I=\int\text{arccosh}(\sqrt{x^2 + 1}) dx=x \text{arcsinh }x-\sqrt{1+x^2}+k$$

For $$x>1$$ $$I=\int arccosh(\sqrt{x^2 + 1})\,dx$$

Using integration by parts;

$$I=arccosh(\sqrt{x^2 + 1})\int \,dx-\int\frac{1}{\sqrt{x^2+1}}\,\left(\int \, dx \,\right)dx$$

$$I=arccosh(\sqrt{x^2 + 1})x-\int\frac{x}{\sqrt{x^2+1}}dx$$ Multiply and divide by 2 in the integral and using U-substitution for $$U=\sqrt{x^2+1}$$;

$$I=arccosh(\sqrt{x^2 + 1})x-\int\frac{x}{\sqrt{x^2+1}}dx$$

$$I=arccosh(\sqrt{x^2 + 1})x-\sqrt{x^2+1}+c$$

Integration by parts $$u=\operatorname{arcosh}(\sqrt{x^2+1}\longrightarrow du=x/|x|\sqrt{x^2+1}dx$$

$$dv=dx \longrightarrow u=x$$

Then;$$\int(udv) = uv-\int(vdu)$$

• For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Commented Jul 15, 2023 at 14:07