Solve Integration: $\int\frac{x}{x^4+x^2+1}dx$

[semc]

\int\frac{x}{x^4+x^2+1}dx

Okay the question ask to integrate the above expression so the first thing i did was to equate u=x² so u'=2x and got this \frac{1}{2}\int\frac{1}{u^2+u+1}du. I then proceed to use integration by parts which i find that this method is too long and messy so i was wondering is there a better to do this

 

[dharavsolanki]

Perfectly well done. But now, convert the ^2 + u + 1 into a friendlier term -->

(u + 0.5)^2 + 0.75

Now, integrate this, it's a direct formula based problem --> will go into tan inverse.

 

[semc]

Damn why didn't i thought of that haha i get it now thanks mate~

 

[dharavsolanki]

If you are solving integrals for the first time, it's really okay.

Try solving the problem with x² in the numerator instead of x. This one is not a simple one, and has a whole lot of pages in the miscellaneous section devoted to it.

 

[semc]

I just notice something, if we are using the anti derivative, shouldn't the denominator be 1+x²? In this case its 0.75 so how do we do it?

 

[dharavsolanki]

Bracket out the 0.75^2 from the denominator and take it outside the integral. Now you have something sqared plus 1 in the denominator. Now, sbstitte that something = v and integrate. Then resubstitute everything, back again.