What is the value of integral $\int \frac{{d}x}{\sqrt {x^2 + 9}}$

Question

What is the value of integral $\int \frac{{d}x}{\sqrt {x^2 + 9}}$
The answer is $\sinh^{-1} (\frac{x}{3})$

I have tried solving it by putting $x = 3\tan \theta$, and got the answer $\ln |sec tan^{-1} \frac{x}{3} + \frac{x}{3}|$

I am unable to convert it in the given answer form, please tell me how to do it.

Answer 1

Do the substitution $x=3\sinh t$ and $\mathrm dx=3\cosh t\,\mathrm dt$. Then your primitive becomes$$\int\frac{3\cosh t}{\sqrt{9\sinh^2(t)+9}}\,\mathrm dt.\tag1$$Since $\sinh^2(t)+1=\cosh^2(t)$, $(1)$ becomes $\int1\,\mathrm dt=t=\operatorname{arcsinh}\left(\frac x3\right)$.

Answer 2

using the substitution $u=x/3$ you get $$\int\frac{dx}{\sqrt{x^2+9}}=\frac{1}{3}\int\frac{dx}{\sqrt{(x/3)^2+1}}=\int\frac{du}{\sqrt{u^2+1}}=\text{arcsinh} (u)=\text{arcsinh}(x/3)$$

Answer 3

As so often happens with questions like these, "different" results are equivalent thanks to a trigonometric identity. On the one hand, $x=3\tan\theta$ does give the result you think it does. On the other hand, $x=3\sinh\phi$ gives the result you've seen somewhere else. You can connect these results viz. $\theta=\operatorname{gd}\phi$.

Answer 4

You want to show that

$$\text{arsinh }x=\ln(x+\sec(\arctan x)).$$

Notice that

$$\sec(\arctan x)=\sqrt{1+\tan^2(\arctan x)}=\sqrt{1+x^2}$$

and that

$$\sinh(\ln t)=\frac12\left(t-\frac1t\right)=\frac{t^2-1}{2t}.$$

So

$$x=\sinh(\text{arsinh }x)=\frac{(x+\sqrt{1+x^2})^2-1}{2(x+\sqrt{1+x^2})}=\frac{x^2+2x\sqrt{1+x^2}+1+x^2-1}{2(x+\sqrt{1+x^2})}$$ is indeed an indentity.