# How to evaluate $\int {{x^2-1}\over{(x^4+3x^2+1)\arctan(\frac{x^2+1}x )}}$ [duplicate]

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.

## marked as duplicate by achille hui calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 10 '16 at 14:54

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.

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