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Integration of ln

Yankel

Active member
Jan 27, 2012
398
Hello all,

I am trying to solve this integral,

\[\int \ln(x^{2}-1) \, dx\]

but I get stuck no matter what I do, if I go for substitution or parts...

thanks
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Here's a hint: $x^2-1=(x+1)(x-1)$.
 

Yankel

Active member
Jan 27, 2012
398
Ok, I tried your suggestion, and I did this, however, it is the wrong answer...

\[\int ln((x-1)(x+1))dx=\int (ln(x-1)+ln(x+1))dx=\]

\[=\int ln(x-1)dx+\int ln(x+1)dx=\]

\[\int ln(u)du+\int ln(t)dt=(x-1)ln(x-1)+(x+1)ln(x+1)-2x+C\]

The correct answer should be:

\[xln(x^{2}-1)-2x-ln\left | x-1 \right |+ln\left | x+1 \right |+C\]
 
Last edited:

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Ok, I tried your suggestion, and I did this, however, it is the wrong answer...

\[\int ln((x-1)(x+1))dx=\int (ln(x-1)+ln(x+1))dx=\]

\[=\int ln(x-1)dx+\int ln(x+1)dx=\]

\[\int ln(u)du+\int ln(t)dt=(x-1)ln(x-1)+(x+1)ln(x+1)-2x+C\]

The correct answer should be:

\[xln(x^{2}-1)-2x-ln\left | x-1 \right |+ln\left | x+1 \right |+C\]
I would approach it in the following manner: first apply integration by parts to get

\[\begin{aligned} \int \ln(x^2-1)\,dx &= x\ln(x^2-1) - \int\frac{2x^2}{x^2-1}\,dx\\ &= x\ln(x^2-1)-\int 2+\frac{2}{x^2-1}\,dx\end{aligned}\]

and then use the hint Ackbach gave you.

Can you take things from here? (Smile)
 

Yankel

Active member
Jan 27, 2012
398
Integration by parts and then use the method of fractions ?

still, I am curious to know what I did wrong earlier...
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Integration by parts and then use the method of fractions ?

still, I am curious to know what I did wrong earlier...
The problem is in claiming that $\ln(x^2-1) = \ln(x-1)+\ln(x+1)$. The property $\ln(ab)=\ln(a)+\ln(b)$ can only be applied when $a>0$ and $b>0$! However, we see that the domain of $\ln(x^2-1)$ is $(-\infty,-1)\cup(1,\infty)$ but the domain of $\ln(x-1)+\ln(x+1)$ is $(1,\infty)$, so we see now that they're not necessarily the "same function"; it's just a portion of $\ln(x^2-1)$. In fact,

\[\ln(x^2-1) = \begin{cases} \ln(x-1)+\ln(x+1) & x>1\\ \ln(1-x) + \ln(-x-1)& x<-1\end{cases}\]

In essence, what you found was $\displaystyle\int \ln(x^2-1)\,dx$ under the constraint that $x>1$. You didn't find $\displaystyle\int \ln(x^2-1)\,dx$ such that it was defined for any $x\in(-\infty,-1)\cup(1,\infty)$. Approaching it by parts and then partial fractions will get you the antiderivative that the solution is looking for.

I hope this made sense (I'm rather tired, so I'm not sure my explanation/reasoning was good enough).
 

Yankel

Active member
Jan 27, 2012
398
Thank you for your help, I understand my mistake.

I tried the way you suggested, and I got something very similar to the answer I look for, however one item is missing.

My solution is attached, my final answer is in bold. Underneath, in red, the real answer according to the book and maple.

I can't find my current mistake either...

Capture.PNG
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Your mistake is in the partial fraction decomposition. It is not true that
$$ \frac{1}{2(x-1)}- \frac{1}{2(x+1)}= \frac{x^{2}}{(x-1)(x+1)}.$$
Just get the common denominator and check it out. Instead, you have
$$\frac{1}{2(x-1)}- \frac{1}{2(x+1)}= \frac{1}{(x-1)(x+1)}.$$
You have to perform polynomial long division first, as Chris mentioned, before you can do the partial fraction decomposition. Indeed, your equation $x^2=A(x+1)+B(x-1)$ should have clued you in that there was a problem: there are no $A$ and $B$ that can make the equation work, because on the LHS you have a quadratic, but no quadratic powers on the RHS.