Compute : $\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$

$$I = \int\frac{x+2}{\sqrt{x^2+5x}+6}\, dx$$ $$\int\frac{\frac{1}{2}(2x+4)}{\sqrt{x^2+5x}+6}\, dx$$ $$\frac{1}{2}\int\frac{2x+ 5 -1}{\sqrt{x^2+5x}+6}\, dx$$ $$\underbrace{\frac{1}{2}\int\frac{2x+5}{\sqrt{x^2+5x}+6}\, dx}_{I_1}\, -\underbrace{\frac{1}{2}\int\frac{1}{\sqrt{x^2+5x}+6}\, dx}_{I_2}$$
The first integral $I_1$ $$I_1 = \frac{1}{2}\int\frac{2x+5}{\sqrt{x^2+5x}+6}\, dx$$ Set $u = x^2+5x \implies \frac{du}{dx} = 2x + 5 \iff dx = \frac{du}{(2x + 5)}$ $$\frac{1}{2}\int\frac{1}{\sqrt{u}+6}\, du $$ Set $t =\sqrt{u} \implies \frac{dt}{du} = \frac{1}{2\sqrt{u}} = \frac{1}{2t} \iff du = 2t\,dt$ $$\int\frac{t}{t + 6}\, dt = \int\left(1 - \frac{6}{t + 6}\right) \, dt = t - 6\log(t +6) + C$$ Substitute back $t = \sqrt{u} = \sqrt{x^2+5x}$

$$I_1 = \sqrt{x^2+5x} - 6\log(\sqrt{x^2+5x} + 6) + C$$

The second integral $I_2$ $$I_2 = -\frac{1}{2}\int\frac{1}{\sqrt{x^2+5x}+6}\, dx$$ $$ -\frac{1}{2}\int\frac{1}{\sqrt{x^2+5x}+6} \,\frac{\sqrt{x^2+5x}-6}{\sqrt{x^2+5x}-6}\, dx$$ $$-\frac{1}{2}\int\frac{\sqrt{x^2+5x}-6}{x^2+5x -36}\, dx$$ $$\underbrace{-\frac{1}{2}\int\frac{\sqrt{x^2+5x}}{x^2+5x -36}\, dx}_{i_1} + \underbrace{3\int\frac{1}{x^2+5x -36}\, dx}_{i_2}$$ We solved $i_2$ $$3\int\frac{1}{x^2+5x -36}\, dx $$ $$3\int\frac{1}{(x+9)(x-4)}\, dx$$ $$3\int\frac{1}{13(x-4)}-\frac{1}{13(x+9)}\, dx$$ $$\frac{3}{13}\left( \int\frac{1}{x-4}\, dx\ - \int\frac{1}{x+9}\, dx\right) $$

$$i_2 = \frac{3}{13}\left( \log(x-4) - \log(x+9) \right) +C$$

We solved $i_1$ $$-\frac{1}{2}\int\frac{\sqrt{x^2+5x}}{x^2+5x -36}\, dx$$ Let $x^2+5x = \left(x + \frac{5}{2} \right)^2 - \frac{25}{4}$ then $$-\frac{1}{2}\int\frac{\sqrt{\left(x + \frac{5}{2} \right)^2 - \left(\frac{5}{2} \right)^2}}{\left(x + \frac{5}{2} \right)^2 - \frac{169}{4}}\, dx = $$ we defeined $a = \frac{5}{2}$ and $u = x + \frac{5}{2} \implies dx = du $ $$-\frac{1}{2}\int\frac{\sqrt{u^2 - a^2}}{u^2 - \left( \frac{13}{2} \right)^2 }\, du$$ By Trigonometric Substitution

$$\text{Let }\sqrt{u^2 - a^2} = \sqrt{a\sec^2\varTheta - a^2} = \sqrt{a^2(\sec^2\varTheta - 1)} = \sqrt{a^2\tan^2\varTheta} = a\tan\varTheta $$ $$\therefore\, u = a\sec\varTheta \implies du = a\sec\varTheta \tan\varTheta d\varTheta$$ With these substitutions, Thus the integral takes the form $$-\frac{1}{2}\int\frac{a\tan\varTheta}{\left( a\sec\varTheta\right) ^2 - \left( \frac{13}{2} \right)^2}\, a\sec\varTheta \tan\varTheta\, d\varTheta$$ Since $\sec(x) = \frac{1}{\cos(x)}$ and $\tan(x) = \frac{\sin(x)}{\cos(x)}$ then we have $$u = a\frac{1}{\cos} \implies du = a\frac{\sin\varTheta}{\cos^2\varTheta} d\varTheta$$ $$-\frac{1}{2}\int\frac{\sqrt{\left( a\frac{1}{\cos\varTheta}\right)^2 - a^2}}{\left( a\frac{1}{\cos\varTheta}\right) ^2 - \left( \frac{13}{2} \right)^2 }\, a\frac{\sin\varTheta}{\cos^2\varTheta} d\varTheta$$ $$-\frac{a}{2}\int\frac{\sin\varTheta\sqrt{ a^2\left(\frac{1}{\cos\varTheta}\right)^2 - a^2}}{\cos^2\varTheta\left( a^2\left(\frac{1}{\cos\varTheta}\right)^2 - \left( \frac{13}{2} \right)^2 \right) }\, d\varTheta$$ $$-\frac{a^2}{2}\int\frac{\sin\varTheta\sqrt{\frac{1}{\cos^2\varTheta} - 1 }}{\cos^2\varTheta\left( a^2\frac{1}{\cos^2\varTheta} - \left( \frac{13}{2} \right)^2 \right) }\, d\varTheta$$ Back $a = \frac{5}{2}$ substitution $$-\frac{25}{8}\int\frac{\sin\varTheta\sqrt{\frac{1}{\cos^2\varTheta} - 1 }}{\cos^2\varTheta\left( \frac{25}{4}\frac{1}{\cos^2\varTheta} - \frac{169}{4} \right) }\, d\varTheta = -\frac{25}{8}\int\frac{\sin\varTheta\sqrt{\frac{1}{\cos^2\varTheta} - 1 }}{\cos^2\varTheta\frac{1}{4}\left(\frac{25}{\cos^2\varTheta} - 169 \right) }\, d\varTheta = -\frac{25}{8}\int\frac{\sin\varTheta\sqrt{\frac{1}{\cos^2\varTheta} - 1 }}{\cos^2\varTheta\frac{1}{4\cos^2\varTheta}\left(25 - 169\cos^2\varTheta \right) }\, d\varTheta$$ $$-\frac{100}{8}\int\frac{\sin\varTheta\sqrt{\frac{1}{\cos^2\varTheta} - 1 }}{25 - 169\cos^2\varTheta }\, d\varTheta$$ Since $\cos^2(x) = \frac{\sin^2(x)}{\tan^2(x)}$ $$-\frac{100}{8}\int\frac{\sin\varTheta\sqrt{\frac{1}{\cos^2\varTheta} - 1 }}{\sin^2\varTheta\left( \frac{25}{\sin^2\varTheta} - \frac{169}{\tan^2\varTheta}\right) }\, d\varTheta$$ Let $\tan(x) = \frac{\sin(x)}{\cos(x)} \implies \tan^2(x) + 1 = \frac{\sin^2(x)}{\cos^2(x)} +1 = \frac{\sin^2(x) + \cos^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)}$ then $\frac{1}{\cos^2(x)} = \tan(x) +1$ and $\frac{1}{tan^2(x) +1} = \frac{1}{\sin^2(x)}$ we have $$-\frac{100}{8}\int\frac{\sin\varTheta\sqrt{\tan^2\varTheta}}{\sin^2\varTheta\left( \frac{25}{\sin^2\varTheta} - \frac{169}{\sin^2\varTheta}+ 169\right) }\, d\varTheta$$ $$-\frac{100}{8}\int\frac{\sin\varTheta\tan\varTheta}{\sin^2\varTheta\left( \frac{-144}{\sin^2\varTheta} + 169\right) }\, d\varTheta$$ Let $\tan(x) = \frac{\sin(x)}{\cos(x)}$ $$-\frac{100}{8}\int\frac{\sin^2\varTheta\frac{1}{\cos\varTheta}}{-\sin^2\varTheta\left( \frac{144}{\sin^2\varTheta} - 169\right) }\, d\varTheta$$ $$\frac{100}{8}\int\frac{\frac{1}{\cos\varTheta}}{\left( \frac{144}{\sin^2\varTheta} - 169\right) }\, d\varTheta$$ $$\frac{100}{8}\int\frac{1}{\frac{\cos\varTheta}{\sin^2\varTheta} \left( 144 - 169\frac{1}{\sin^2\varTheta}\right) }\, d\varTheta$$ Set $$w = \frac{1}{\sin \varTheta} \implies dw = -\frac{\cos\vartheta}{\sin^2\varTheta}$$ $$-\frac{100}{8}\int\frac{1}{144 - 169w^2 }\, wd$$ $$-\frac{100}{8}\int\frac{1}{(12 + 13w)(12 - 13w)}\, wd$$ $$-\frac{100}{8}\int\frac{1}{24(12 + 13w)}-\frac{1}{24(13w - 12)}\, wd$$ $$-\frac{25}{48}\left( \int\frac{1}{12 + 13w}\, wd-\int\frac{1}{13w - 12}\, wd\right) $$ $$-\frac{25}{48}\left( \log(12 + 13w) - \log(13w - 12)\right) +C$$ Now, back substitute to get $$u = \frac{5}{2}\frac{1}{\cos\varTheta} \implies \cos\varTheta = \frac{5}{2u}$$ Let $\sin(x) = \sqrt{1- \cos^2(x)}$ $$\sin\varTheta = \sqrt{1- \cos^2\varTheta} = \sqrt{1- \left( \frac{5}{2u}\right) ^2} = \frac{1}{2u}\sqrt{4u^2 - 25}$$ $$w = \frac{1}{\sin \varTheta} = \frac{2u}{\sqrt{4u^2 - 25}}$$ $$u = x + \frac{5}{2}$$ $$w = \frac{2\left( x + \frac{5}{2}\right) }{\sqrt{4\left( x + \frac{5}{2}\right)^2 - 25}} = \frac{2x + 5}{\sqrt{4(x^2 + 5x + \frac{25}{4}) - 25} } = \frac{2x + 5}{2\sqrt{x^2 + 5x}}$$ Also

$$i_1 = -\frac{25}{48}\left( \log\left( 12 + 13\frac{2x + 5}{2\sqrt{x^2 + 5x}}\right) - \log\left( 13\frac{2x + 5}{2\sqrt{x^2 + 5x}} - 12\right) \right) +C$$

The Solution: $$ I = I_1 + i_1 + i_2$$ Set $t = x^2 + x5$

$$I = \sqrt{t} - 6\log(\sqrt{t} + 6) -\frac{25}{48}\left( \log\left( 12 + 13\frac{2x + 5}{2\sqrt{t}}\right) - \log\left( 13\frac{2x + 5}{2\sqrt{t}} - 12\right) \right) + \frac{3}{13}\left( \log(x-4) - \log(x+9) \right) + C$$

$$ \int \frac{\sqrt{x^2+5x}}{x^2+5x-36}~dx$$

$$=\int \frac{\sqrt{x^2+5x}+6-6}{(\sqrt{x^2+5x})^2-6^2}~dx$$

$=\int \frac{\sqrt{x^2+5x}+6}{(\sqrt{x^2+5x}+6)(\sqrt{x^2+5x}-6)} - \int \frac{6}{x^2+5x-36} ~dx$

$$=\int \frac{1}{\sqrt{x^2+5x}-6} - \int \frac{6}{x^2+5x-36} ~dx$$

Hope you can proceed further.

Introduce the Euler substitution:

Let $u=\dfrac{\sqrt{x^2+5x}}{x}$ ,

Then $x=\dfrac{5}{u^2-1}$

$dx=-\dfrac{10u}{(u^2-1)^2}~du$

$\therefore\int\dfrac{x+2}{\sqrt{x^2+5x}+6}~dx$

$=-\int\dfrac{\dfrac{5}{u^2-1}+2}{\dfrac{5u}{u^2-1}+6}\dfrac{10u}{(u^2-1)^2}~du$

$=-\int\dfrac{(2u^2+3)10u}{(6u^2+5u-6)(u^2-1)^2}~du$

$=-\int\dfrac{(2u^2+3)10u}{(3u-2)(2u+3)(u+1)^2(u-1)^2}~du$

$=\int\left(-\dfrac{5}{2(u+1)^2}-\dfrac{5}{2(u-1)^2}+\dfrac{11}{2(u+1)}+\dfrac{13}{2(u-1)}-\dfrac{144}{13(2u+3)}-\dfrac{252}{13(3u-2)}\right)~du$ (according to http://www.wolframalpha.com/input/?i=-((2u%5E2%2B3)10u)%2F((3u-2)(2u%2B3)(u%2B1)%5E2(u-1)%5E2))

$=\dfrac{5}{2(u+1)}+\dfrac{5}{2(u-1)}+\dfrac{11\ln(u+1)}{2}+\dfrac{13\ln(u-1)}{2}-\dfrac{72\ln(2u+3)}{13}-\dfrac{84\ln(3u-2)}{13}+C$

$=\dfrac{5u}{u^2-1}+\dfrac{11\ln(u+1)}{2}+\dfrac{13\ln(u-1)}{2}-\dfrac{72\ln(2u+3)}{13}-\dfrac{84\ln(3u-2)}{13}+C$

$=\sqrt{x^2+5x}+\dfrac{11}{2}\ln\dfrac{x+\sqrt{x^2+5x}}{x}+\dfrac{13}{2}\ln\dfrac{\sqrt{x^2+5x}-x}{x}-\dfrac{72}{13}\ln\dfrac{3x+2\sqrt{x^2+5x}}{x}-\dfrac{84}{13}\ln\dfrac{3\sqrt{x^2+5x}-2x}{x}+C$