Evaluate the following definite integral $$\int_{0}^{1} \frac{x}{(x^2+x+4)\sqrt{4x^2+4x+5}}\,dx.$$
My steps:
Separating the integral gives $$\int_{0}^{1}\left(\frac{8x+4}{8(x^2+x+4)\sqrt{4x^2+4x+5}}-\frac1{2(x^2+x+4)\sqrt{4x^2+4x+5}}\right)dx.$$ For the first integral, letting $$u=\sqrt{4x^2+4x+5}$$ $$du=\frac{8x+4}{2\sqrt{4x^2+4x+5}}\ dx$$ gives $$\int_{0}^{1}\frac{8x+4}{8(x^2+x+4)\sqrt{4x^2+4x+5}}\ dx = \frac14\int_{\sqrt5}^{\sqrt{13}} \frac1{x^2+x+4}du$$ $$$$Knowing that $$x^2+x+4=\frac{u^2+11}4$$ The integral in terms of $u$ becomes: $$\int_{\sqrt5}^{\sqrt{13}}\frac1{u^2+11}\ du=\frac{\sqrt{11}}{11}\int_{\sqrt5}^{\sqrt{13}}\frac1{\sqrt{11}\left(\left(\frac{u}{\sqrt{11}}\right)^2+1\right)}\,du$$ Evaluating the indefinite integral, we get: $$\frac1{\sqrt{11}}\arctan\left(\sqrt{\frac{4x^2+4x+5}{11}}\right)+C$$ For the second integral, let:$$u=\frac{8x+4}{\sqrt{4x^2+4x+5}}$$ $$du=\frac{32}{\sqrt{(4x^2+4x+5)^3}}\ dx$$ Thus: $$-\int_0^1\frac1{2(x^2+x+4)\sqrt{4x^2+4x+5}}\,dx=-\frac1{64}\int_{4/\sqrt5}^{12/\sqrt{13}}\frac{4x^2+4x+5}{x^2+x+4}\,du$$ To write the integrand in terms of $u$, we need to find values for $a$ and $b$ such that: $$\frac{x^2+x+4}{4x^2+4x+5}=a\cdot u^2+b\Leftrightarrow\frac{x^2+x+4}{4x^2+4x+5}=a\cdot\frac{64x^2+64x+16}{4x^2+4x+5}+b\cdot\frac{4x^2+4x+5}{4x^2+4x+5}$$ $$\Leftrightarrow x^2+x+4=(64a+4b)x^2+(64a+4b)x+(16a+5b)$$ $$$$By solving a system of linear equations for $a$ and $b$ for the relations: $$\begin{cases} 64a+4b=1 \\ 16a+5b=4 \end{cases}$$ we find that: $$ a=-\frac{11}{256}\quad \wedge \quad b=\frac{15}{16}.$$ Therefore, the integral in terms of $u$ becomes: $$-\frac1{64}\int_{4/\sqrt5}^{12/\sqrt{13}}\frac{4x^2+4x+5}{x^2+x+4}\ du=-4\int_{4/\sqrt5}^{12/\sqrt{13}}\frac{1}{240-11u^2}\,du=$$ $$=-\frac1{60}\int_{4/\sqrt5}^{12/\sqrt{13}}\frac{1}{1-\left(\frac{\sqrt{11}u}{4\sqrt{15}}\right)^2}\,du$$ Letting: $$s=\frac{\sqrt{11}}{4\sqrt{15}}u$$ $$ds=\frac{\sqrt{11}}{4\sqrt{15}}du$$ and subsequently substituting for $s$, obtaining: $$\displaystyle-\frac1{\sqrt{165}}\int_{\sqrt{11/75}}^{\sqrt{33/65}}\frac{1}{1-s^2}\ ds$$ $$$$Evaluating the indefinite integral, we get: $$-\frac1{\sqrt{165}}\tanh^{-1}\left(\frac{\sqrt{11}(2x+1)}{\sqrt{15(4x^2+4x+5)}}\right)+C$$ $$$$Now we can evaluate the integral from $0$ to $1$: $$\int_{0}^{1} \frac{x}{(x^2+x+4)\sqrt{4x^2+4x+5}}\ dx=$$ $$=\left[\frac1{\sqrt{11}}\arctan\left(\sqrt{\frac{4x^2+4x+5}{11}}\right)-\frac1{\sqrt{165}}\tanh^{-1}\left(\frac{\sqrt{11}(2x+1)}{\sqrt{15(4x^2+4x+5)}}\right)\right]_0^1=$$ $$\boxed{\frac1{\sqrt{11}}\arctan\left(\sqrt{\frac{13}{11}}\right)-\frac1{\sqrt{165}}\tanh^{-1}\left(\sqrt{\frac{33}{65}}\right)-\\\frac1{\sqrt{11}}\arctan\left(\sqrt{\frac{5}{11}}\right)+\frac1{\sqrt{165}}\tanh^{-1}\left(\sqrt{\frac{11}{75}}\right)}$$
Question: Is this procedure correct? Also, is there a quicker way to evaluate the integral?