Definite integration. $$\int _0^1 \arctan(x^2-x+1)\,dx$$
ATTEMPT:
$\int \arctan(x^2-x+1)\,dx$
Let
$\arctan(x^2-x+1)=u$ and $dx=dv$: 
$$du=\frac{2x−1}{(x^2−x+1)^2+1}dx= \frac{2x−1}{(x^2+1)(x^2-2x+2)}$$ and $v=x$.
Now from Integration by parts $I=uv-$$\int vdu.$
$$I=\arctan(x^2-x+1)x-\int \frac{x(2x−1)}{(x^2+1)(x^2-2x+2)}.$$
Now using partial fractions
$$
\int\frac{x(2x−1)}{(x^2+1)(x^2-2x+2)}=\int \frac{x}{(x^2+1)}-\frac{x}{x^2-2x+2} $$
Integrating term by term and finally arranging the result:
$$I=x\arctan(x^2−x+1)+\frac{1}{2}\ln(∣x^2+1∣)−\frac{1}{2}\ln(∣x^2−2x+2∣)−\arctan(x−1).$$
Now substituting the limit i got $I=\ln2.$
Is there any other way to approach this problem like by using the properties of definite integration as this method is long and tedious but though it works!!

This response is similar in spirit to @nospoon's answer, but I think it is perhaps more elementary because it doesn't require the anti-derivative of arctangent.
$$\begin{align}
I
&=\int_{0}^{1}\arctan{\left(x^2-x+1\right)}\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\frac{\pi}{2}--\arctan{\left(\frac{1}{x^2-x+1}\right)}\right]\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\frac{\pi}{2}-\arctan{\left(x\right)}-\arctan{\left(1-x\right)}\right]\,\mathrm{d}x\\
&=\frac{\pi}{2}\int_{0}^{1}\mathrm{d}x-\int_{0}^{1}\arctan{\left(x\right)}\,\mathrm{d}x-\int_{0}^{1}\arctan{\left(1-x\right)}\,\mathrm{d}x\\
&=\frac{\pi}{2}\int_{0}^{1}\mathrm{d}x-2\int_{0}^{1}\arctan{\left(x\right)}\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\frac{\pi}{2}-2\arctan{\left(x\right)}\right]\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\operatorname{arccot}{\left(x\right)}-\arctan{\left(x\right)}\right]\,\mathrm{d}x\\
&=\int_{0}^{1}\frac{2x}{1+x^2}\,\mathrm{d}x\\
&=\ln{\left(1+x^2\right)}\bigg{|}_{0}^{1}\\
&=\ln{(2)}.\\
\end{align}$$
