I am pretty sure that my answer is correct but given answer for the exercise from textbook Calculus James Steward was slightly different. Any idea to solve this:
$$\int\frac{x}{\sqrt{x^2+x+1}} \, dx$$
The given answers: $\sqrt{x^2+x+1}-\frac{1}{2}\ln(\sqrt{x^2+x+1}+x+\frac{1}{2})+c$
Thanks in advance.
On
First split off the bit that does not need a trigonometric substitution: $$\int\frac{x\,dx}{\sqrt{x^2+x+1}}=\int\frac{x+\frac{1}{2}}{\sqrt{x^2+x+1}}dx -\frac{1}{2}\int\frac{dx}{\sqrt{x^2+x+1}}\ .$$ You should see that the first integral is now easy. For the second, complete the square: $$\int\frac{dx}{\sqrt{x^2+x+1}}=\int\frac{dx}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}}\ .$$ If you really want to use a trig substitution, try $$x+\frac{1}{2}=\frac{\sqrt3}{2}\tan\theta\ ,$$ which still leaves you with a slightly tricky integral. Better, use the hyperbolic substitution $$x+\frac{1}{2}=\frac{\sqrt3}{2}\sinh\theta\ .$$ If you want to directly get the answer from the book, do some algebra: $$\frac{1}{\sqrt{x^2+x+1}} =\frac{\displaystyle\frac{x+\frac{1}{2}}{\sqrt{x^2+x+1}}+1}{\sqrt{x^2+x+1}+x+\frac{1}{2}}\ .$$ With a bit of thought you will see why this is easy to integrate.
$$x^2+x+1 = \left ( x+\frac12 \right )^2+\frac{3}{4} $$
$$\begin{align} \int dx \frac{x}{\sqrt{x^2+x+1}} &= \int dx \frac{x+\frac12}{\sqrt{\left ( x+\frac12 \right )^2+\frac{3}{4}}}-\frac12 \int dx \frac{1}{\sqrt{\left ( x+\frac12 \right )^2+\frac{3}{4}}}\end{align}$$