Integrate $\frac{1}{\sqrt{x^2+cx}}$

Question

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.

Answer 1

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 $C =C'-\ln|2c|+\ln(2)$

Answer 2

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.

Answer 3

Substitute $\sqrt{x^2+cx}=\left(x+\frac c2\right)t$ instead, or $\left(x+\frac c2\right)^2=\frac{c^2}{4(1-t^2)}$. Then $$dt = \frac{c^2}{4\left(x+\frac c2\right)^2 \sqrt{x^2+cx}}dx =\frac{1-t^2}{\sqrt{x^2+cx}}dx $$ and $$\int \frac1{\sqrt{x^2+cx}}dx=\int \frac1{1-t^2}dt=\tanh^{-1}t =\tanh^{-1}\frac{\sqrt{x^2+cx}}{x+\frac c2} $$