Find $\int \sinh^{-1}x\hspace{1mm}dx$ Find $\int \sinh^{-1}x\hspace{1mm}dx$
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I am asked to use the following Equation:
$$\int \tan^{-1}x\hspace{1mm}dx= x\tan^{-1}x-\ln(\sec(\tan^{-1}x))+C$$
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The confusing part is : What has $\sinh^{-1}x$ to do with $\tan^{-1}x$ 
Any Ideas
 A: The idea is to use integration by parts and substitution. For example, using the substitution $x=\sinh(u)$, we get
$$
\begin{align}
\int\sinh^{-1}(x)\,\mathrm{d}x
&=\int u\,\mathrm{d}x\\
&=xu-\int x\,\mathrm{d}u\\
&=x\sinh^{-1}(x)-\int\sinh(u)\,\mathrm{d}u\\[3pt]
\end{align}
$$
which is a basic integral for hyperbolic functions. If you don't know that integral, remember that
$$
\sinh(x)=\frac{e^x-e^{-x}}{2}\qquad\text{and}\qquad\cosh(x)=\frac{e^x+e^{-x}}{2}
$$
A: $\sinh^{-1}(x)$ has nothing to do with $\tan^{-1}(x)$, it doesn't hint you to find a correlation between them, it hints you to use the same method for solving the integral. This is the dormant hint in there;
$$\int f \mathrm{d}g = fg-\int g \mathrm{d}f$$
$$f=\sinh^{-1}(x), \mathrm{d}f=\frac{1}{\sqrt{x^2+1}} \mathrm{d}x, g=x, \mathrm{d}g=\mathrm{d}x$$ which makes your integral $$\int \sinh^{-1}(x) \mathrm{d}x= x \cdot \sinh^{-1}(x) -\int \frac{x}{\sqrt{x^2+1}} \mathrm{d}x$$
I believe you can take it from here.
A: This is what $\sinh^{-1}x$ has to do with $\tan^{-1} x$.
$$\frac{1}{x^2+1} = \bigg(\frac{1}{\sqrt{x^2+1}}\bigg)^2$$
$$\frac{d}{dx}\tan^{-1}x = \bigg(\frac{d}{dx}\sinh^{-1}x\bigg)^2$$
This fact (that both derivatives are based on $x^2+1$) comes from the similarity in the following identities
$$\sec^2 x=\tan^2x +1$$
$$\cosh^2 x = \sinh^2 x + 1$$
