Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$

Question

Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$

I solved this integral by euler substitution by replacing

$\sqrt{x^2+x-1}=x+t$

but it's not allowed by the problem.

p.s Is there any other method to solve with?

Thank you in advance :)

Answer 1

$$ I = \int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx = \int\frac{3x^2-1}{\sqrt{(x + \frac12)^2 - \frac54}}dx $$ Replacing $x= y\cdot \sqrt{\frac54} - \frac12$ gives $$ I = \int\frac{(15 y^2)/4 - (3 \sqrt 5 y)/2 - 1/4}{\sqrt{(y^2-1)}}dy $$ Now let $y = \cosh (z)$ which gives $$ I = \int [15 (\cosh (z))^2)/4 - (3 \sqrt 5 \cosh (z))/2 - 1/4 ] \; dz $$ This can easily be performed, giving $$I = \frac{13 z - 12 \sqrt 5 \sinh(z) + 15 \sinh(z) \cosh(z)}{8} + {\rm constant}$$ Replacing $z$ by $x$ (via $y$) gives $$ I = \frac18 \Big[6 \sqrt{x^2 + x - 1} (2 x - 3) + 13 \tanh^{-1}\frac{2 x + 1}{2 \sqrt{x^2 + x - 1}} \Big] + {\rm constant} $$

Answer 2

Integrate by parts to get

\begin{align} \int \sqrt{x^2+x-1} \>dx &\overset{ibp}=\frac12 x \sqrt{x^2+x-1} +\frac14\int \frac {x-2}{\sqrt{x^2+x-1}}dx\\ \end{align} and then use it below to evaluate \begin{align} &\int\frac{3x^2-1}{\sqrt{x^2+x-1}}\>dx\\ =& \ 3 \int \sqrt{x^2+x-1} \ dx - \int \frac {3x-2}{\sqrt{x^2+x-1}}\ dx\\ =& \ \frac32 x \sqrt{x^2+x-1} -\frac94\int \frac {x+\frac12}{\sqrt{x^2+x-1}}dx + \frac{13}8 \int \frac {1}{\sqrt{x^2+x-1}}dx \\ =& \ \left( \frac32 x -\frac94 \right)\sqrt{x^2+x-1} + \frac{13}8 \coth^{-1}\frac{x+\frac12}{\sqrt{x^2+x-1}}+C \end{align}

Answer 3

Hint:

We observe that $$p':=\left(\sqrt{x^2+x-1}\right)'=\frac{2x+1}{2\sqrt{x^2+x-1}},$$ $$q':=\left(x{\sqrt{x^2+x-1}}\right)'=\frac{4x^2+3x-2}{2\sqrt{x^2+x-1}}.$$

Also, by completing the square,

$$x^2+x-1=\left(x+\frac12\right)^2-\frac54=\frac54\left(\left(\frac{2x+1}{\sqrt5}\right)^2-1\right)$$

and

$$r':=\left(\text{arcosh}\frac{2x+1}{\sqrt5}\right)'=\frac1{\sqrt{x^2+x-1}}$$

Hence we form a linear combination to match the numerator of the integrand and we find

$$\frac{3q'}2-\frac{9p'}4+\frac{13r'}8=\frac{3x^2-1}{\sqrt{x^2+x-1}}.$$