MHB Solving the Integral of sqrt{x^2+1} Using Substitution

[karush]

I know this was done before on this forum but can't find it

$\displaystyle \int\sqrt{x^2+1}\ dx%$

$\text{where $u=x$ and $a=1$ then plug}$

$\displaystyle \frac{u}{2}\sqrt{u^2+a^2} + \frac{a^2}{2}\ln\left| u+\sqrt{u^2+a^2}\right|$

how is this derived?

 

[MarkFL]

Re: int sqrt{x^2+1}

Let's begin with:

$$I=\int\sqrt{u^2+a^2}\,du$$

Now, consider the hyperbolic trig. identity:

$$\cosh^2(x)=\sinh^2(x)+1$$

Hence, for some positive real constant $a$, we have:

$$a^2\cosh^2(x)=a^2\sinh^2(x)+a^2$$

And so if we use:

$$u=a\sinh(x)\implies du=a\cosh(x)$$

The integral becomes:

$$I=a^2\int \cosh^2(x)\,dx$$

Next, consider the identiy:

$$\cosh^2(x)=\frac{\cosh(2x)+1}{2}$$

And our integral becomes:

$$I=\frac{a^2}{2}\int \cosh(2x)+1\,dx=\frac{a^2}{2}\left(\frac{1}{2}\sinh(2x)+x\right)+C$$

Next, consider the identity:

$$\sinh(2x)=2\sinh(x)\cosh(x)$$

and an implication of a previously used identity:

$$\cosh(x)=\sqrt{\sinh^2(x)+1}$$

And we have:

$$I=\frac{a^2}{2}\left(\sinh(x)\sqrt{\sinh^2(x)+1}+x\right)+C$$

Back-substitute for $x$:

$$I=\frac{a^2}{2}\left(\frac{u}{a}\sqrt{\left(\frac{u}{a}\right)^2+1}+\arsinh\left(\frac{u}{a}\right)\right)+C$$

$$I=\frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}\arsinh\left(\frac{u}{a}\right)+C$$

Now, consider the identity:

$$\arsinh(x)=\ln\left(x+\sqrt{x^2+1}\right)$$

And we have:

$$I=\frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}\ln\left(\frac{u}{a}+\sqrt{\left(\frac{u}{a}\right)^2+1}\right)+C$$

$$I=\frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}\ln\left(u+\sqrt{u^2+a^2}\right)-\frac{a^2}{2}\ln(a)+C$$

Since $$\frac{a^2}{2}\ln(a)$$ is a constant, we may write:

$$I=\frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}\ln\left(u+\sqrt{u^2+a^2}\right)+C$$

 

[karush]

very helpful

I can see now why its a plug in equation

i tried it for a few steps but I lost the trail thru the woods