How can I solve $\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$

Question

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?

Answer 1

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 $$ t=\tanh \frac u2, \quad \sinh u= \frac{2t}{1-t^2},\quad du=\frac{2}{1-t^2}\:dt. $$

Answer 2

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 $$\int -\frac{2 (t-7) t(2t-7)}{9\left(21-t+t^2\right)}dt$$

Answer 3

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: $$\int\frac{1}{\sqrt{u^2-a^2}+3}=\color{green}{\log\left(\sqrt{u^2-a^2}+u\right)+}\color{green}{\frac{3\times coth^{-1}\left(\frac{v}{\sqrt{a^2+9}}\right)}{\sqrt{a^2+9}}}\color{green}{-\frac{3\times coth^{-1}\left(\frac{u}{\sqrt{a^2+9}}\right)}{\sqrt{a^2+9}}}+C$$