Помогите взять интеграл от √(x^2+9) Пожалуйста...

Решите задачу:

$\int \sqrt{x^2+9}\, dx=[\; x=3tgt\; ,\; dx= \frac{3\, dt}{cos^2t} \; ,\; t=arctg\frac{x}{3}\; ,=\\\\ \=x^2+9=9tg^2t+9=9(tg^2+1)=9\cdot \frac{1}{cos^2t} \, ]=\int \frac{\frac{3}{cost}\cdot 3\, dt}{cos^2t} =\\\\=9\, \int \frac{dt}{cos^3t} =9\, \int \frac{sin^2t+cos^2t}{cos^3t}\, dt=9\, \int \frac{sin^2t\, dt}{cos^3t}+9\, \int \frac{dt}{cost}=A\; ;\\\\ \int \frac{sint\cdot sint\, dt}{cos^3t}=[\, u=sint\; ,\; du=cost\, dt\; ,\; dv= \frac{sint\, dt}{cos^3t} \; ,\\\\v=-\int \frac{-sint\, dt}{cos^3t}=-\int \frac{d(cost)}{cos^3t}=- \frac{(cost)^{-2}}{-2} = \frac{1}{2cos^2t} \; ]=$

$= \frac{sint}{2cos^2t}-\int \frac{cost\, dt}{2cos^2t} =\frac{sint}{2cos^2t}- \frac{1}{2} \int \frac{dt}{cost} \; ;\\\\A=\frac{9\, sint}{2cos^2t}-\frac{9}{2}\int \frac{dt}{cost} +9\int \frac{dt}{cost} = \frac{9\, sint}{2cos^2t}+\frac{9}{2}\int \frac{dt}{cost} \; ;\\\\\star \; \; \int \frac{dt}{cost} =\int \frac{cost\, dt}{cos^2t}=\int \frac{cost\, dt}{1-sin^2t} =\int \frac{d(sint)}{1-sin^2t}=\frac{1}{2}\, ln\Big |\frac{1+sint}{1-sint} \Big |+C\; \star \\\\A= \frac{9sint}{2cos^2t}+\frac{9}{4}\, ln\Big | \frac{1+sint}{1-sint} \Big |+C\; ,\; t=arctg\frac{x}{3}\; .$