# Math Help - integrating ln

1. ## integrating ln

I'm trying to integrate $\int\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx$
$u = \ln(x+\sqrt{1+x^2})$
$\frac{du}{dx}=\frac{1}{x}(1+\frac{x}{\sqrt{1+x^2}} )=\frac{1}{x}+\frac{1}{\sqrt{1+x^2}}$

Can I do like this? $\ln{x} + \ln{\sqrt{1+x^2}}=\ln(x+\sqrt{1+x^2})$
if so, then do I continue like this? $\int{2u} du= u^2$

Hm... and I noticed that I rewrited $\ln$ backward xD omg, can anyone help me out here?

2. A clue here is the $\sqrt{1 + x^2}$ ... if that x were a tan of something, how would that change things?

Just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case theta), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

The general drift is...

Spoiler:

Late edit: Soroban's solution below is a bit simpler, as it does without the trigonometry. Just in case a picture helps...

Spoiler:

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

3. Originally Posted by Revy
I'm trying to integrate $\int\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx$
$u = \ln(x+\sqrt{1+x^2})$
$\frac{du}{dx}=\frac{1}{x}(1+\frac{x}{\sqrt{1+x^2}} )=\frac{1}{x}+\frac{1}{\sqrt{1+x^2}}$

Can I do like this? $\ln{x} + \ln{\sqrt{1+x^2}}=\ln(x+\sqrt{1+x^2})$
if so, then do I continue like this? $\int{2u} du= u^2$

Hm... and I noticed that I rewrited $\ln$ backward xD omg, can anyone help me out here?
No, the logarithm of a sum is not the sum of the logarithms! You should review the basic properties of logarithms.

Here's a hint : the numerator looks very much like the expressions one gets for the inverse hyperbolic functions (such as arcsinh, arctanh, etc).

4. Hello, Revy!

I got the answer . . . but it was a strange journey.

$\displaystyle{\int\frac{\ln(x+\sqrt{1+x^2})}{\sqrt {1+x^2}}\,dx$

We have: . $\displaystyle \int \ln(x+\sqrt{1+x^2})\cdot\frac{dx}{\sqrt{1+x^2}}$ .[1]

I used the same substitution, but I found a sneaky step.

Let $u \,=\,x + \sqrt{1+x^2}$ .[2]

Then: . $du \:=\:\left(1 + \dfrac{x}{\sqrt{1+x^2}}\right)\,dx \quad\Rightarrow\quad du\:=\:\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}\,dx$

From [2], we have: . $du \:=\:\dfrac{u}{\sqrt{1+x^2}}\,dx \quad\Rightarrow\quad \dfrac{dx}{\sqrt{1+x^2}} \,=\,\dfrac{du}{u}$ .[3]

Substitute [2] and [3] into [1]: . $\displaystyle \int \ln u\cdot\frac{du}{u}$

Let $v \,=\,\ln u \quad\Rightarrow\quad dv \,=\,\frac{du}{u}$

Substitute: . $\displaystyle{\int v\,dv \;=\;\tfrac{1}{2}v^2 + C$

Back-substitute: . $\frac{1}{2}[\ln u]^2 + C$

Back-substitute: . $\frac{1}{2}\bigg[\ln\!\left(x+\sqrt{1+x^2}\,\right)\bigg]^2 + C$

5. The substitution $u=ln\left(x+\sqrt{1+x^2}\right)$ will solve it directly.

6. Could you explain me this part? $\dfrac{dx}{\sqrt{1+x^2}} \,=\,\dfrac{du}{u}$

According to book answer should be $\ln^2\!\left(x+\sqrt{1+x^2}\,\right) + C
$
but I see that more than one person got the same answer, then there might be mistake in the book

Originally Posted by Ted
The substitution $u=ln\left(x+\sqrt{1+x^2}\right)$ will solve it directly.
If I substitute whole ln then it goes like that?
$du=\frac{1}{x+\sqrt{1+x^2}} \frac{x+\sqrt{1+x^2}{\sqrt{1+x^2}}}dx=\frac{dx}{\s qrt{1+x^2}}$
and then as was posted before
$\int {\ln{u}du}$
$v=\ln u \quad\Rightarrow\quad dv =\frac{du}{u}$
$\displaystyle{\int v\,dv \;=\;\tfrac{1}{2}v^2 + C$

7. Originally Posted by Revy
I'm trying to integrate $\int\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx$

$u = \ln(x+\sqrt{1+x^2})$

correct

$\frac{du}{dx}=\frac{1}{x}(1+\frac{x}{\sqrt{1+x^2}} )=\frac{1}{x}+\frac{1}{\sqrt{1+x^2}}$

No! the $\displaystyle\;\;\frac{1}{x}\;\;$ term should be $\diplaystyle\;\;\frac{1}{x+\sqrt{1+x^2}}$

Think of it as

$u=lny\Rightarrow\ \frac{du}{dx}=\frac{du}{dy}\ \frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}$

Can I do like this?

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

This breaks the laws of logarithms.

if so, then do I continue like this? $\int{2u} du= u^2$

Hm... and I noticed that I rewrited $\ln$ backward xD omg, can anyone help me out here?

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

$\displaystyle\frac{du}{dx}=\frac{1}{y}\ \frac{dy}{dx}=\frac{1}{x+\sqrt{1+x^2}}\left(1+\fra c{2x}{2\sqrt{1+x^2}}\right)$

$=\displaystyle\frac{1}{x+\sqrt{1+x^2}}\left(1+\fra c{x}{\sqrt{1+x^2}}\right)=\frac{1}{x+\sqrt{1+x^2}} \left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right)$

$=\displaystyle\frac{1}{\sqrt{1+x^2}}$

$\Rightarrow\int{u}du=\frac{u^2}{2}+C$

since...

$\displaystyle\frac{du}{dx}=\frac{1}{\sqrt{1+x^2}}\ Rightarrow\ du=\frac{dx}{\sqrt{1+x^2}}\Rightarrow\int{ln\left( x+\sqrt{1+x^2}\right)\left(\frac{dx}{\sqrt{1+x^2}} \right)=\int{u}du$