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?

  • $\begingroup$ If the numerical solution is not the same as your own result, you must have made a mistake somewhere $\endgroup$
    – Yuriy S
    Feb 1, 2018 at 22:28
  • $\begingroup$ @YuriyS NIntegrate agrees with mine. But this is not the point, the main question is if there is a shorter way to this. $\endgroup$
    – Denis28
    Feb 1, 2018 at 22:30
  • 1
    $\begingroup$ For your second question, Euler substitution might lead to a faster solution, though I can't say for sure until I try. See here en.m.wikipedia.org/wiki/Euler_substitution $\endgroup$
    – Yuriy S
    Feb 1, 2018 at 22:51

2 Answers 2


Sometimes an integral involving a quadratic polynomial can be simplified a bit by letting $u$ equal to a multiple of the derivative of the quadratic.

This particular case involves two quadratics with derivatives $2x+1$ and $4(2x+1)$ suggesting a substitution $u=2x+1$ resulting in the slight reduction in the original:

$$ \int_1^3\frac{u-1}{(u^2+15)\sqrt{u^2+4}}\,du $$

  • $\begingroup$ I think $u=2 \sinh z$ might be a good substitution here $\endgroup$
    – Yuriy S
    Feb 2, 2018 at 0:14

The Euler substitution method doesn't seem to lead to a shorter solution, but still, it's quite nice.

First, we make substitution:


Now we find:

$$x=\frac{1}{4} \frac{5-t^2}{(t-1)}$$

And after some tedious algebra, the integral becomes:

$$2 \int_{\sqrt{5}}^{\sqrt{13}-2} \frac{(t^2-5)dt}{\left((t-1)^2+4 \right)^2+44(t-1)^2}$$

Now we make another substitution:


And we obtain:

$$\int_{(\sqrt{5}-1)/2}^{(\sqrt{13}-3)/2} \frac{(p^2+p-1)dp}{\left(p^2+1 \right)^2+11p^2}$$

Fortunately, the denominator can be factored over real numbers, so we obtain:

$$\int_{(\sqrt{5}-1)/2}^{(\sqrt{13}-3)/2} \frac{(p^2+p-1)dp}{\left(p^2+\frac{13}{2}- \frac{\sqrt{165}}{2}\right)\left(p^2+\frac{13}{2}+ \frac{\sqrt{165}}{2}\right)}$$

From here we use partial fractions as usual.

The advantage of Euler substitution and other such methods is that it's algorithmic. You can use it for any cases where the expression is a rational function of a single radical of a quadratic function. So there's no need to guess.


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