How to Solve an Integral/Diff Problem: Finding Antiderivative of (x^2)/(4-(x^2))

[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.

 

[amcavoy]

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.

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

Then use partial fractions.

 

[Muzly]

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

Whoops - my bad!

 

[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.