Integrate

**charikaar** · December 6th 2009, 09:30 AM

what is the easiest way to integrate this

$\frac{-14x+18x^3}{(3x^2-1)(1-x^2)}$

thanks

**VonNemo19** · December 6th 2009, 09:55 AM

Originally Posted by charikaar

what is the easiest way to integrate this

By partial fraction decompostion.

**charikaar** · December 6th 2009, 10:03 AM

is this correct

$\frac{-14x+18x^3}{ (\sqrt{3}x - 1)(\sqrt{3}x+1)(1+x)(1-x)} = \frac{2\sqrt3}{\sqrt{3}x - 1}+\frac{-2\sqrt3}{\sqrt{3}x + 1} + \frac{1}{1-x} + \frac{-1}{1+x}$

**Amer** · December 6th 2009, 10:03 AM

Originally Posted by charikaar

what is the easiest way to integrate this

$\frac{-14x+18x^3}{(3x^2-1)(1-x^2)}$

thanks

I do not know if my way is the easiest but here it is

$\frac{18x^3 -14x }{(3x^2-1)(1-x^2)}$

$\left(\frac{18x^3-14x}{(3x^2-1)}\right)\left(\frac{1}{1-x^2}\right)$

$\left(6x - \frac{8x}{(3x^2-1)} \right) \left(\frac{1}{1-x^2} \right)$

$\int \frac{6x}{1-x^2 } - \frac{8x}{(3x^2-1)(1-x^2)} \cdot dx$

sub $x^2 = u \Rightarrow 2x \cdot dx = du \Rightarrow dx = \frac{du}{2x}$

$\int \frac{3}{1-u} - \frac{4}{(3u-1)(1-u)} \cdot du$

second one you can integrate it by partial fraction first one is easy

**VonNemo19** · December 6th 2009, 10:44 AM

Originally Posted by charikaar

is this correct

$\frac{-14x+18x^3}{ (\sqrt{3}x - 1)(\sqrt{3}x+1)(1+x)(1-x)} = \frac{2\sqrt3}{\sqrt{3}x - 1}+\frac{-2\sqrt3}{\sqrt{3}x + 1} + \frac{1}{1-x} + \frac{-1}{1+x}$

I tried to check it for you, but I think that there is an easier way.

Note that

$\frac{18x^3-14x}{(3x^2-1)(x-1)(x+1)}=\frac{Ax+B}{3x^2-1}+\frac{C}{x-1}+\frac{D}{x+1}$

$18x^3-14x=(Ax+B)(x^2-1)+C(3x^2-1)(x+1)+D(3x^2-1)(x-1)$

$x=1:$ $4=4C\Rightarrow{C}=1$

$x=-1:$ $-4=-4D\Rightarrow{D}=1$

Now With the values of C and D known, we put them in and then

$x=0:$ $0=-B-1+1\Rightarrow{B}=0$

And now use any arbitrary value of x to solve for A.

$x=2\Rightarrow{A}=12$ s.t.

$\frac{18x^3-14x}{(3x^2-1)(x-1)(x+1)}=\frac{12x}{3x^2-1}+\frac{1}{x-1}+\frac{1}{x+1}$

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