# Integral/Diff Problem

1. Oct 31, 2005

### Muzly

OK, here's the question:

Differentiate x.ln(4-(x^2))
Which is:

ln(4-(x^2)) - (2(x^2))/(4-(x^2))

Now, find an antiderivative of (x^2)/(4-(x^2)).

Answer says -x + ln( (2+x)/(2-x) ) +c

How do I get that? The comments say "Many failed to recognise the need to divide" - What does that mean?

I'm so damn stuck! Please help! Any is appreciated.

2. Oct 31, 2005

### amcavoy

$$\int\frac{x^2}{4-x^2}\,dx=\int\frac{x^2}{\left(2-x\right)\left(2+x\right)}\,dx$$

Then use partial fractions.

3. Oct 31, 2005

### Muzly

Well aren't I a bit silly - should have long-divided then used partial fractions!!!

Whoops - my bad!

4. Oct 31, 2005

### Benny

Muzly - Are you doing the VCE?

I don't think you can use partial fractions(well you can use it but it won't get you anywhere) if the degree of the numerator is not strictly less than the degree of the denominator.

Use the following and then use integration by recognition to complete the question.

$$\frac{{x^2 }}{{4 - x^2 }} = \frac{{ - \left( {4 - x^2 - 4} \right)}}{{4 - x^2 }} = - 1 + \frac{4}{{4 - x^2 }}$$

Long divison is not really needed for this question.

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