# Thread: Integration by parts

1. ## Integration by parts

$\displaystyle \int ln(x^2-4)dx=xln(x^2-4)-\int \frac{x(2x)}{x^2-4}dx$

$\displaystyle =xln(x^2-4)-(xln(x^2-4)+\int ln(x^2-4)dx$

$\displaystyle =\int ln(x^2-4)dx$

Where did i go wrong?

2. ## Re: Integration by parts

Originally Posted by Punch
$\displaystyle \int ln(x^2-4)dx=xln(x^2-4)-\int \frac{x(2x)}{x^2-4}dx$

$\displaystyle =xln(x^2-4)-(xln(x^2-4)+\int ln(x^2-4)dx$

$\displaystyle =\int ln(x^2-4)dx$

Where did i go wrong?
note that $\displaystyle \ln(x^2-4) = \ln(x+2) + \ln(x-2)$

I would break the original integrand into the two terms above, then use integration by parts for each integral.

3. ## Re: Integration by parts

Originally Posted by Punch
$\displaystyle \int ln(x^2-4)dx=xln(x^2-4)-\int \frac{x(2x)}{x^2-4}dx$<--------------You did right till here.

$\displaystyle =xln(x^2-4)-(xln(x^2-4)+\int ln(x^2-4)dx$

$\displaystyle =\int ln(x^2-4)dx$

Where did i go wrong?
...

You have $\displaystyle I=x\ln{(x^2-4)}-\int \frac{2x^2}{x^2-4}dx$

\displaystyle \begin{align*} &=x\ln{(x^2-4)}-2\int \frac{x^2-4+4}{x^2-4}dx = x\ln{(x^2-4)}-2{\left[ x+\int \frac{4}{x^2-4}dx\right]} \end{align*}

$\displaystyle = x\ln{(x^2-4)}-2{\left[ x+\int \frac{1}{x-2}dx-\int\frac{1}{x+2}dx\right]}$

Proceed

4. ## Re: Integration by parts

Originally Posted by sbhatnagar
...

You have $\displaystyle I=x\ln{(x^2-4)}-\int \frac{2x^2}{x^2-4}dx$

\displaystyle \begin{align*} &=x\ln{(x^2-4)}-2\int \frac{x^2-4+4}{x^2-4}dx = x\ln{(x^2-4)}-2{\left[ x+\int \frac{4}{x^2-4}dx\right]} \end{align*}

$\displaystyle = x\ln{(x^2-4)}-2{\left[ x+\int \frac{1}{x-2}dx-\int\frac{1}{x+2}dx\right]}$

Proceed
Thank you, I realised something, integrating an improper fraction is a NO-NO