Integral using Euler's substitution Question:
Solve this:
$$\displaystyle \int \frac {2x-\sqrt {4x^2-x+1} } {x-1} \,dx$$
Our solution: 
$$16 \ln|t+4|-\ln|t+0.25|+C$$ when $t=\sqrt {4x^2-x+1}-2x$ (using Euler's formula)
But wolfram's solution was totally different. We were wondering if these two are equal.
We found this  question very helpful
Integral( using Euler substitution)
but we were wondering how could one check that hyperbolic solution and Euler solution are equal?
 A: Tools that might help for a direct check if the two solutions are equal up to a constant ... and perhaps useful in the future.


*

*We have that the expression $\sqrt{4x^2-x+1}-2x$ can be re-written as $\frac{(\sqrt{4x^2-x+1}-2x)(\sqrt{4x^2-x+1}+2x)}{\sqrt{4x^2-x+1}-2x}=\frac{(4x^2-x+1)-4x^2}{\sqrt{4x^2-x+1}+2x}$. 

*Moreover, if $\frac{(4x^2-x+1)-4x^2}{\sqrt{4x^2-x+1}+2x}$ is inside a logarithm it can be re-written as $\log(1-x)-\log(\sqrt{4x^2-x+1}+2x)$.

*The inverse of $\sinh$ can be computed in terms of logarithms: $y:=\sinh(x)=\frac{e^x-e^{-x}}{2}$, from where $e^{2x}-2ye^x-1=0$, and then $e^x=\frac{2y+\sqrt{4y^2+4}}{2}$. Then we get $\sinh^{-1}(x)=\log(y+\sqrt{y^2+1})$.
A: Two solutions of a given indefinite integral may differ only by some constany $c$. So to check if the solutions are equal you should subtract one solution from the other and check if it's constant.
A: Never forget that there is a very basic rule to check if your answer is correct: compute its derivative! If you get the function under the integral sign, you are done.
Of course the situation is different if you want to check directly that two primitives differ by a constant. In this case, there is no general rule, and you must use particular tricks that depend on the form of the two primitives. For instance, the inverse function of sinh and cosh can be expressed in terms of logarithms, by solving explicitly the equation $\sinh x = y$ or $\cosh x =y$.
A: This is the same integral as the other question, since
$$\frac{2x - \sqrt{4x^2-x+1}}{x-1}=\frac{2x - \sqrt{4x^2-x+1}}{4x^2 - (4x^2-x+1)}=\frac{1}{2x+\sqrt{4x^2-x+1}}$$
