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How do I go about doing this: $\displaystyle \int {{x^2-1}\over{(x^4+3x^2+1)\arctan(\frac{x^2+1}x )}}$

I have tried integration by parts but it seems to be making the problem more complicated.

Also, I have thought of substitution but I couldn't find any suitable substitution.

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marked as duplicate by achille hui calculus Nov 10 '16 at 14:54

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Notice, substitute $\text{u}=\text{f}\left(\text{x}\right)$ and $\text{d}\text{u}=\text{f}\space'\left(\text{x}\right)\space\text{d}\text{x}$:

$$\int\frac{\text{f}\space'\left(\text{x}\right)}{\text{f}\left(\text{x}\right)}\space\text{d}\text{x}=\int\frac{1}{\text{u}}\space\text{d}\text{u}=\ln\left|\text{u}\right|+\text{C}=\ln\left|\text{f}\left(\text{x}\right)\right|+\text{C}$$

Now, when:

$$\text{f}\left(\text{x}\right)=\arctan\left\{x+\frac{1}{x}\right\}$$

And:

$$\text{f}\space'\left(\text{x}\right)=\frac{x^2-1}{x^4+3x^2+1}$$


So, when you want to prove the result:

$$\int\frac{\frac{x^2-1}{x^4+3x^2+1}}{\arctan\left\{x+\frac{1}{x}\right\}}\space\text{d}\text{x}=\int\frac{x^2-1}{\left(x^4+3x^2+1\right)\arctan\left\{x+\frac{1}{x}\right\}}\space\text{d}\text{x}=\ln\left|\arctan\left\{x+\frac{1}{x}\right\}\right|+\text{C}$$

But look at this.

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  • $\begingroup$ @MrYouMath Thank you very much!!! $\endgroup$ – Jan Nov 10 '16 at 15:00

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