Integrating $\frac{1}{\sqrt{x^2+1}}$ with the help of a variable change

Question

The question I am faced with is as follows:

With the help of a variable change $\sqrt{x^2+1}=x+t$, demonstrate that over $\mathbb{R}$, with $k\in \mathbb{R}$

$$\int{\frac{1}{\sqrt{x^2+1}} \ \mathrm{d}}x=\ln(x+\sqrt{x^2+1})+k$$

Now what I did was something different, I built a right triangle with an angle $\alpha$ (its adjacent side being $1$ and opposite side being $x$), and with a hypotenuse of $\sqrt{x^2+1}$.

Thus, $x=\tan{\alpha}$ and $\mathrm{d}x=\sec^2{\alpha} \ \mathrm{d}\alpha$. So we have:

$$\int{\frac{1}{\sqrt{x^2+1}} \ \mathrm{d}}x=\int{\cos{\alpha} \ \mathrm{d}x}=\int{\frac{\sec^2{\alpha}}{\sec{\alpha}} \ \mathrm{d}\alpha}=\int{\sec{\alpha} \ \mathrm{d}\alpha}$$

I know how to integrate $\sec{\alpha} \ \mathrm{d}\alpha$, and after replacing $\alpha$ with the initial value of $x$, I indeed end up with the result $\ln(x+\sqrt{x^2+1})+k$.

But how would one end up with the same result using the given hint? Can someone help me out? Thanks.

Answer 1

Let $t=\sqrt{x^2+1}-x=\frac{1}{\sqrt{x^2+1}+1}$, which is equivalent to $x=\frac{1-t^2}{2t}$. Then, we have

$$\begin{align} \int \color{blue}{\frac{1}{\sqrt{1+x^2}}}\,\color{red}{dx}&=\int \color{blue}{\left(\frac{2t}{1+t^2}\right)}\,\color{red}{\left(-\frac12 \frac{t^2+1}{t^2}\right)\,dt}\\\\ &=-\int \frac1t\,dt\\\\ &=-\log(t)+k\\\\ &=\log(x+\sqrt{x^2+1})+k \end{align}$$

as was to be shown!

Answer 2

With $\sqrt{x^2+1}=t-x$ (slightly better) you have $x^2+1=t^2-2tx+x^2$, so $$ x=\frac{t^2-1}{2t}=\frac{t}{2}-\frac{1}{2t} $$ so also $$ dx=\frac{1}{2}\left(1+\frac{1}{t^2}\right)\,dt=\frac{t^2+1}{2t^2}\,dt $$ and $$ \sqrt{x^2+1}=t-x=t-\frac{t}{2}+\frac{1}{2t}=\frac{t^2+1}{2t} $$ Then the integral becomes $$ \int\frac{2t}{t^2+1}\frac{t^2+1}{2t^2}\,dt $$