Trouble with the integral Help

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Trouble with the integral! Help

While I was working some examples of integrals from book, I found this example:

\int {{x^3-1}\over{4x^3+x}}
and I can not solve it, because I get always incorrect result.

Result is: (from book)

{x \over 4}+{{x \over 16}*{ln ({{x^{16}}\over{(2x-1)^7(2x+1)^9}}})}

Can you help me solve it!

Thanks.
 
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Before you do anything with the integral, take polynomial long division, since the degree of the numerator >= degree of the denominator.
 


depecheSoul said:
While I was working some examples of integrals from book, I found this example:

\int {{x^3-1}\over{4x^3+x}}
and I can not solve it, because I get always incorrect result.

Result is: (from book)

{x \over 4}+{{x \over 16}*{ln ({{x^{16}}\over{(2x-1)^7(2x+1)^9}}})}

Can you help me solve it!

Thanks.

You can notice x^3 - 1 is of the form "a^3-b^3". You can factorize ;) !

And you can factorize: 4x^3 + x.

It is a partial fraction.

Good luck ;) !
 


Thank you!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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