Question

Помогите решить неопределенные интегралы

Answer 1

$1)\; \; \int \frac{dx}{\sqrt{4-9x^2}}=\int \frac{dx}{\sqrt{2^2-(3x)^2}}=\frac{1}{3}\int \frac{3\cdot dx}{\sqrt{2^2-(3x)^2}}=[\, t=3x,\; dt=3\, dx\, ]=\\\\=\frac{1}{3}\int \frac{dt}{\sqrt{2^2-t^2}}=\frac{1}{3}\cdot arcsin\frac{t}{2}+C=\frac{1}{3}\cdot arcsin \frac{3x}{2}+C\; ;\\\\2)\; \; \int \frac{dx}{(x+2)\cdot ln(x+2)}=\int\frac{\frac{dx}{x+2}}{ln(x+2)}=[\, t=ln(x+2),\; dt=\frac{dx}{x+2}\, ]=\\\\=\int \frac{dt}{t}=ln|t|+C=ln|ln(x+2)|+C\; ;$

$3)\; \; \int \frac{x\cdot dx}{cos^2(3x^2+5)}=\frac{1}{6}\int \frac{6x\cdot dx}{cos^2(3x^2+5)}=[\, t=3x^2+5,\; dt=6x\, dx\, ]=\\\\=\frac{1}{6}\int \frac{dt}{cos^2t}=\frac{1}{6}\cdot tgt+C=\frac{1}{6}\cdot tg(3x^2+5)+C\; ;\\\\4)\; \; \int \frac{dx}{x^2+6x+7}=\int \frac{dx}{(x+3)^2-2}=[\, t=x+3,\; dt=dx\, ]=\int \frac{dt}{t^2-(\sqrt2)^2}=\\\\=\frac{1}{2\sqrt2}\cdot ln\Big |\frac{t-\sqrt2}{t+\sqrt2}\Big |+C=\frac{1}{2\sqrt2}\cdot ln\Big |\frac{x+3-\sqrt2}{x+3+\sqrt2}\Big |+C\; .$