How can I solve $\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$ How can I solve the following integral?
 $$\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$$
$$\begin{align}
\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx&=\int \frac{1}{\sqrt{\left(x+\frac72\right)^2-\frac{49}{4}} + 3}\, dx
\end{align}$$
$$u = x+\frac{7}{2}, \quad a = \frac{7}{2}$$
$$\int \frac{1}{\sqrt{u^2 - a^2} +3}\, \,du$$
Attempt I - By Trigonometric substitution
$$\sqrt{u^2 - a^2} = \sqrt{a\sec^2\varTheta - a^2} = \sqrt{a^2(\sec^2\varTheta - 1)} = \sqrt{a^2\tan^2\varTheta} = a\tan\varTheta $$
$$u = a\sec\varTheta \implies du = a\sec\varTheta \tan\varTheta d\varTheta$$
 $$\int \frac{1}{a\tan\varTheta + 3}\, a\sec\varTheta \tan\varTheta d\varTheta$$
 $$a\int \frac{\sec\varTheta \tan\varTheta}{a\tan\varTheta + 3}\, d\varTheta$$
How can I continue here ?
Attempt II - By First Euler Substitution
$$\sqrt{u^2 - a^2} = u - t$$
$\sqrt{u^2 - a^2} = u - t$
$ u^2 - a^2 = (u - t)^2$
$u^2 - a^2 = u^2 -2ut + t^2$
$- a^2 = -2ut + t^2$
$2ut =  t^2 + a^2$
$$u =  \frac{t^2 + a^2}{2t} \implies du = \frac{1}{2} - \frac{a^2}{2t^2}\, dt$$
Thus the integral takes the form
$$\int \frac{1}{u - t +3}\, \left( \frac{1}{2} - \frac{a^2}{2t^2}\right) \,du$$
Since $u =  \frac{t^2 + a^2}{2t}$ and $\left( \frac{1}{2} - \frac{a^2}{2t^2}\right)  = -\frac{a^2 - t^2}{2t^2}$ then
$$I = \int \frac{1}{ \frac{t^2 + a^2}{2t}  - t +3} \left( -\frac{a^2 - t^2}{2t^2} \right) \, dt$$
$$-\int \frac{a^2 - t^2}{ ta^2 - t^3 +6t^2}\, dt$$
$$-\int \frac{a^2}{ ta^2 - t^3 +6t^2}\, dt + \int \frac{t^2}{ ta^2 - t^3 +6t^2}\, dt$$
How can I continue?
Attempt III -By First Euler Substitution
$$\sqrt{x^2 + 5x} = x + t$$
$\sqrt{x^2 + 5x} = x + t$
$ x^2 + 5x = (x + t)^2$
$x^2 + 5x = x^2 + 2ut + t^2$
$x^2 + 5x = x^2 + 2xt + t^2$
$5x = 2xt + t^2$
$5x -2xt = t^2$
$x(5 -2t) = t^2$
$$x = \frac{t^2}{(5 -2t)} \implies \frac{dt}{dx} = -\frac{2(t - 5)t}{(5 - 2t)^2 } \iff dx =  -\frac{(5 - 2t)^2 }{2(t - 5)t}\,dt $$
$$\int \frac{1}{x + t + 3}\, \left( -\frac{(5 - 2t)^2 }{2(t - 5)t}\right)dt$$ 
$$-\int \frac{1}{\frac{t^2}{(5 -2t)} + t + 3}\, \left(\frac{(5 - 2t)^2 }{2(t - 5)t}\right)dt$$ 
$$-\int\frac{(5 - 2t)^2}{\frac{2(t - 5)t^3}{5 -2t} + 2(t - 5)t^2 + 6(t - 5)t}\,dt$$
How can I continue here?
 A: Setting $$\sqrt{x^2+7x}=t+x$$ then we have $$x=\frac{t^2}{7-2t}$$ and $$dx=-\frac{2 (t-7) t}{(2 t-7)^2}$$ and $$t+x=\frac{(t-7) t}{2 t-7}$$
and our integral will be $$\int -\frac{2 (t-7) t(2t-7)}{9\left(21-t+t^2\right)}dt$$
A: Start from your first result and multiply numerator and denominator with $\sqrt{u^2-a^2}-3$:
$$\int\frac{1\times \color{red}{\sqrt{u^2-a^2}-3}}{\sqrt{u^2-a^2}+3 \times \color{red}{\sqrt{u^2-a^2}-3}}du=\int\frac{\sqrt{u^2-a^2}-3}{u^2-a^2-9}du$$
Now we are going to split this integration into two parts. Each part will be calculated separately:
$$\int\color{blue}{\frac{\sqrt{u^2-a^2}}{u^2-a^2-9}du} -\int\color{purple}{\frac{3}{u^2-a^2-9}du}$$
First we calculate the blue part of the integration. We do this by multiplying numerator and denominator with $\sqrt{u^2-a^2}$. Then the integral can be broken as follows:
$$\int\frac{\sqrt{u^2-a^2}\times\color{red}{\sqrt{u^2-a^2}}}{u^2-a^2-9\times\color{red}{\sqrt{u^2-a^2}}}du=\int\frac{u^2-a^2}{(u^2-a^2-9)\sqrt{u^2-a^2}}du$$
$$=\int\frac{A}{\sqrt{u^2-a^2}}du+\int\frac{B}{u^2-a^2-9}du$$
Then the appropriate values of $A$ and $B$ are:
$$A\left(u^2-a^2-9\right)+B\left(\sqrt{u^2-a^2}\right)=u^2-a^2 \implies A=1, B=\frac{9}{\sqrt{u^2-a^2}}$$
The integral becomes:
$$\int\frac{1}{\sqrt{u^2-a^2}}du+\int\frac{\frac{9}{\sqrt{u^2-a^2}}}{u^2-a^2-9}du$$
This first part can be calculated using Euler substitution:
$$\int\frac{1}{\sqrt{u^2-a^2}}=\color{green}{\log\left(\sqrt{u^2-a^2}+u\right)} +C$$
We introduce $9u^2$ in the denominator and we multiply both numerator and denominator with $a^2$. Then we divide this with $u^2-a^2$.
$$\int\frac{\frac{9a^2}{\sqrt{u^2-a^2}}}{\color{red}{9u^2-9u^2}+u^2a^2-a^4-9a^2}du=\int\frac{\frac{9a^2}{(u^2-a^2)^\frac{3}{2}}}{\frac{9u^2}{u^2-a^2}-a^2-9}du$$
Then we introduce a variable $v$:
$$v=\frac{3u}{\sqrt{u^2-a^2}}\implies dv=\frac{3a^2}{(u^2-a^2)^\frac{3}{2}}\implies \int\frac{3}{v^2-a^2-9}dv=\int\frac{3}{v^2-\left(\sqrt{a^2+9}\right)^2}dv=\color{green}{\frac{3\times coth^{-1}\left(\frac{v}{\sqrt{a^2+9}}\right)}{\sqrt{a^2+9}}}+C$$
The purple part of the integration can be calculated identically:
$$\int\frac{3}{u^2-a^2-9}du=\int\frac{3}{u^2-\left(\sqrt{a^2+9}\right)^2}du=\color{green}{\frac{3\times coth^{-1}\left(\frac{u}{\sqrt{a^2+9}}\right)}{\sqrt{a^2+9}}}+C$$Finally:
$$\int\frac{1}{\sqrt{u^2-a^2}+3}=\color{green}{\log\left(\sqrt{u^2-a^2}+u\right)+}\color{green}{\frac{3\times coth^{-1}\left(\frac{v}{\sqrt{a^2+9}}\right)}{\sqrt{a^2+9}}}\color{green}{-\frac{3\times coth^{-1}\left(\frac{u}{\sqrt{a^2+9}}\right)}{\sqrt{a^2+9}}}+C$$
A: Hint. By the change of variable
$$
x+\frac72=\frac72 \cdot \cosh u,\qquad dx=\frac72 \cdot \sinh u\:du,
$$ one gets
$$
\begin{align}
\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx&=\int \frac{1}{\sqrt{\left(x+\frac72\right)^2-\frac{49}{4}} + 3}\, dx
\\\\&=\int \frac{\frac72\sinh u}{\frac72\sinh u + 3}\, du
\end{align}
$$ then one may conclude with the change of variable
$$
t=\tanh \frac u2, \quad \sinh u= \frac{2t}{1-t^2},\quad du=\frac{2}{1-t^2}\:dt.
$$ 
