how to integrate this ?

Question

anonymous · Answer

$$\int\limits \frac{ 1 }{ x^4+3x^2-4 } dx$$

anonymous · Answer

Partial fractions? If possible

anonymous · Answer

Denominator is factorizable I think...

anonymous · Answer

Yep, just checked with Wolfram

anonymous · Answer

show me the steps please

anonymous · Answer

And I checked with my eyes.. :P

ganeshie8 · Answer

have u factored the denominator yet ? it would be good if you show that you're trying instead of asking for a step by step solution... no one would be interested in  doing your homework for you (just saying)

anonymous · Answer

Notice that the denominator is zero if you plug in $x=\pm1$. This means you can use long division to find the last factor(s):
$$\dfrac{x^4+3x^2-4}{x^2-1}=\cdots$$

anonymous · Answer

@ganeshie8 : i am not asking people to do my homework...else i would have posted a list of sums, which i did not.

ganeshie8 · Answer

All good ! keep going :)

aum · Answer

$
x^4+3x^2-4 = x^4 + 4x^2 - x^2 - 4 = x^2(x^2+4) - 1(x^2+4) = \
(x^2-1)(x^2+4) = (x-1)(x+1)(x^2+4)
$
Now try partial fractions.

agent0smith · Answer

If you can factor $$\large x^2 +3x  -4 $$then you can factor $$\large x^4 +3x^2  -4 $$since it's if a quadratic equation, if you make a substitution  and let u = x^2 $$\large u^2 +3u  -4$$

anonymous · Answer

@agent0smith  thank you