Integrate $\frac{1}{\sqrt{x^2+cx}}$ I am trying to compute the integral $\int\frac{1}{\sqrt{x^2+cx}}dx$.
To begin I completed the square of the denominator resulting in $$\int \frac{1}{\sqrt{(x+\frac{c}{2})^2-\frac{c^2}{4}}}dx$$
I then made the substitution $u=x+\frac{c}{2}$, which has left me with the integral $$\int \frac{1}{\sqrt{u^2-\frac{c^2}{4}}}du$$
From here I am stumped on what to do next. Is there a simpler way to integrate this? Am I heading in the right direction? Any help would be appreciated.
 A: Hint:
A table of derivatives reveals that
$$(\text{arcosh}(t))'=\left(\ln\left|t+\sqrt{t^2-1}\right|\right)'=\frac1{\sqrt{t^2-1}}.$$
It takes a simple linear transform to make the connection.
A: You've done it properly.
continuing from where you left off ;
$I =\displaystyle\int \frac{1}{\sqrt{u^2-\frac{c^2}{4}}}du$
$I =\displaystyle\int \frac{1}{\frac{c}2\sqrt{\frac{4u^2}{c^2}-1}}du$
let $ \frac{2u}{c} = \sec(t) \implies  \frac 2c \,du = \sec(t)\tan(t)\,dt$
$I =\displaystyle\int\frac{1}{\sqrt{\sec^2(t)-1}}\sec(t)\tan(t)\,dt$
$I = \displaystyle \int\sec(t)\,dt $
$I = \ln|\sec(t)+\tan(t)|+C'$
$I = \ln\bigg|\frac{2u}c+ \sqrt{\frac{4u^2}{c^2}-1}\bigg|+C'$
$I = \ln\bigg|\frac{2(x+\frac c2)}c+ \sqrt{\frac{4(x+\frac c2)^2}{c^2}-1}\bigg|+C'$
EDIT :
$I =\ln\bigg|\frac{4x+2c}{2c}+ \sqrt{\frac{4(2x+c)^2-4c^2}{4c^2}}\bigg|+C'$
$I =\ln\bigg|\frac{4x+2c}{2c}+ \sqrt{\frac{4(2x+c)^2-4c^2}{4c^2}}\bigg|+C' $
$I= \ln\bigg|\frac{4x+2c}{2c}+ \frac{\sqrt{{4(2x+c)^2-4c^2}}}{2c}\bigg|+C'$
$I= \ln\bigg|2({2x+c}+ {\sqrt{{(2x+c)^2-c^2}}})\bigg|-\ln|2c|+C'$
$I = I= \ln\bigg|({2x+c}+ {\sqrt{{(2x+c)^2-c^2}}}\bigg|+\ln(2)-\ln|2c|+C'$
$I= \ln\bigg|{2x+c}+ {\sqrt{{(2x+c)^2-c^2}}}\bigg|+C$$\quad$ Where $C =C'-\ln|2c|+\ln(2)$
