Help in finding the integrate of a quadratic. $$\int_0^{0.904463} \sqrt{1+(2.343548x-0.812495)^2 }dx$$
I have tried this question several times but can't answer it. I know it involves trig functions a u-subsitution.
 A: $$\begin{align}
\int\sqrt{1+(ax-b)^2}dx&\stackrel{ax-b\mapsto\sinh t}=\frac1a\int\cosh t\; d\sinh t= \frac1a\int\cosh^2 t\; dt\\
&=\frac1{2a}\int(1+\cosh 2t)\; d t= \frac1{2a}\left(t+\frac12\sinh2t\right)\\
&=\frac{t+\sinh t\cosh t}{2a}\\
&=\frac{\sinh^{-1}(ax-b)+(ax-b)\sqrt{1+(ax-b)^2}}{2a}.
\end{align}$$
The substitution of the numbers is left to you.
A: I will answer for the general formula
$$\int\sqrt{x^2+2px+q}\,dx$$
(in your case, you can pull the coefficient of $x$ out of the square root; mind the $2p$).
We can write 
$$x^2+2px+q=(x+p)^2+q-p^2=(x+p)^2\pm d^2.$$
We observe that
$$\left(x\sqrt{x^2+2px+q}\right)'=\sqrt{x^2+2px+q}+\frac{x^2+px}{\sqrt{x^2+2px+q}}
\\=2\sqrt{x^2+2px+q}-\frac{px+q}{\sqrt{x^2+2px+2q}}$$
and
$$\left(\sqrt{x^2+2px+q}\right)'=\frac{x+p}{\sqrt{x^2+2px+q}}$$
and 
$$\left(\text{arsinh}\frac{x+p}d\right)'=\frac1{d\sqrt{(x+p)^2+d^2}}$$ or $$\left(\text{arcosh}\frac{x+p}d\right)'=\frac1{d\sqrt{(x+p)^2-d^2}}.$$
Hence the antiderivative is of the form
$$(ax+b)\sqrt{x^2+2px+q}+c\text{ arsinh/arcosh}\frac{x+p}d$$
where $a,b,c$ can be obtained by the method of indeterminate coefficients.
