Integration by parts

**Punch** · December 9th 2011, 04:31 AM

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

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

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

Where did i go wrong?

**skeeter** · December 9th 2011, 04:38 AM

Originally Posted by Punch

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

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

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

Where did i go wrong?

note that $\ln(x^2-4) = \ln(x+2) + \ln(x-2)$

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

**sbhatnagar** · December 9th 2011, 04:58 AM

Originally Posted by Punch

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

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

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

Where did i go wrong?

...

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

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

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

Proceed

**Punch** · December 9th 2011, 05:38 AM

Originally Posted by sbhatnagar

...

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

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

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

Proceed

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

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