1. ## Indefinite Integral

What is the value of $\displaystyle \int \frac{x+2}{x^2+x+1}dx$?

Can it be solved? If yes, what is the answer?

2. ## Re: Indefinite Integral

Yes it can be solved, but you have to rewrite some things. We want to recognize the first derivative of the denominator which is $\displaystyle 2x+1$ in the numerator, so:
$\displaystyle \int \frac{x+2}{x^2+x+1}dx $$\displaystyle =\frac{1}{2} \int \frac{2x+4}{x^2+x+1}dx=\frac{1}{2}\int \frac{2x+1+3}{x^2+x+1}dx$$\displaystyle =\frac{1}{2}\int \frac{2x+1}{x^2+x+1}dx +\frac{3}{2}\int \frac{dx}{x^2+x+1}$

For the second integral use completing the square in the denominator.

3. ## Re: Indefinite Integral

Have I done correct?

$\displaystyle \int \frac{x+2}{x^2+x+1}dx$

$\displaystyle =\frac{1}{2} \int \frac{2x+1}{x^2+x+1}dx+\frac{3}{2}\int \frac{1}{x^2+x+1}dx$

$\displaystyle =\frac{1}{2} \ln{(x^2+x+1)}+\frac{3}{2}\int \frac{1}{x^2+x+1}dx$

$\displaystyle =\frac{1}{2} \ln{(x^2+x+1)}+\frac{3}{2}\int \frac{1}{(\frac{2x+1}{2})^2+(\frac{\sqrt{3}}{2})^2 }dx$

$\displaystyle =\frac{1}{2} \ln{(x^2+x+1)}+\sqrt{3}\tan^{-1}(\frac{2x+1}{\sqrt{3}})$

4. ## Re: Indefinite Integral

Originally Posted by sbhatnagar
Have I done correct?

$\displaystyle \int \frac{x+2}{x^2+x+1}dx$

$\displaystyle =\frac{1}{2} \int \frac{2x+1}{x^2+x+1}dx+\frac{3}{2}\int \frac{1}{x^2+x+1}dx$

$\displaystyle =\frac{1}{2} \ln{(x^2+x+1)}+\frac{3}{2}\int \frac{1}{x^2+x+1}dx$

$\displaystyle =\frac{1}{2} \ln{(x^2+x+1)}+\frac{3}{2}\int \frac{1}{(\frac{2x+1}{2})^2+(\frac{\sqrt{3}}{2})^2 }dx$

$\displaystyle =\frac{1}{2} \ln{(x^2+x+1)}+\sqrt{3}\tan^{-1}(\frac{2x+1}{\sqrt{3}})$
Why didn't you do as instructed and complete the square in the denominator of the second integral?