Evaluate $\int \frac{1}{(2x+1)\sqrt {x^2+7}}\,\text{d}x$.

Question

How to do this indefinite integral (anti-derivative)?

$$I=\displaystyle\int \dfrac{1}{(2x+1)\sqrt {x^2+7}}\,\text{d}x.$$

I tried doing some substitutions ($x^2+7=t^2$, $2x+1=t$, etc.) but it didn't work out.

Answer 1

Using Euler substitution by setting $t-x=\sqrt{x^2+7}$, we will obtain $x=\dfrac{t^2-7}{2t}$ and $dx=\dfrac{t^2+7}{2t^2}\ dt$, then the integral turns out to be \begin{align} \int \dfrac{1}{(2x+1)\sqrt {x^2+7}}\ dx&=\int\frac{1}{t^2+t-7}\ dt\\ &=\int\frac{1}{\left(t+\dfrac{\sqrt{29}+1}{2}\right)\left(t-\dfrac{\sqrt{29}-1}{2}\right)}\ dt\\ &=-\int\left[\frac{2}{\sqrt{29}(2t+\sqrt{29}+1)}+\frac{2}{\sqrt{29}(-2t+\sqrt{29}-1)}\right]\ dt. \end{align} The rest can be solved by using substitution $u=2t+\sqrt{29}+1$ and $v=-2t+\sqrt{29}-1$.

Answer 2

Use Trigonometric substitution,

$$x=\sqrt7\tan\theta$$

Then $\displaystyle a\sin y+b\cos y=\sqrt{a^2+b^2}\sin\left(y+\arctan \frac ba\right)$ and this

or use Weierstrass substitution

Answer 3

Let $x=\sqrt{7}\tan{u}\implies dx=\sqrt{7}\sec^2{u}du$ \begin{align} I =\int \frac{1}{(2x+1)\sqrt{x^2+7}}dx=\int \frac{\sec{u}}{(1+2\sqrt{7}\tan{u})}du\\ \end{align} Let $t=\tan{\frac{u}{2}}\implies du=\frac{2}{1+t^2}dt$ \begin{align} I =\int\frac{\frac{1+t^2}{1-t^2}}{1+2\sqrt{7}\frac{2t}{1-t^2}}\frac{2}{1+t^2}dt=-2\int \frac{1}{t^2-4\sqrt{7}t-1}dt \end{align} Decompose the integrand into partial fractions and integrate. \begin{align} I=-2A\ln{(t-2\sqrt{7}-\sqrt{29})}-2B\ln{(t-2\sqrt{7}+\sqrt{29})}+c \end{align} where \begin{align} A={\rm Res}(f,2\sqrt{7}+\sqrt{29})=\frac{1}{2\sqrt{29}}, B={\rm Res}(f,2\sqrt{7}-\sqrt{29})=-\frac{1}{2\sqrt{29}} \end{align} Now express everything in terms of x.

Answer 4

Substitute $x=\sqrt7 \tan t$, with $t\in (-\frac\pi2,\frac\pi2)$

\begin{align} &\int \frac{1}{(2x+1)\sqrt {x^2+7}}\ dx \\ =& \int \frac1{\cos t+2\sqrt7\sin t}\ dt =\int \frac{d( {\sin t -2\sqrt7\cos t})}{29-( {\sin t -2\sqrt7\cos t})^2}\\ =&\ \frac1{\sqrt{29}}\tanh^{-1}\frac{\sin t-2\sqrt7\cos t}{\sqrt{29}} = \frac1{\sqrt{29}}\tanh^{-1}\frac{x-14}{\sqrt{29}\sqrt{x^2+7}} \end{align}