Question

Помогите решить а, в, г, д

Answer 1

$1)\; \; \int cos(3x+5)dx=\frac{1}{3}sin(3x+5)+C\; ;\\\\2)\; \; \int (\underbrace {3-x}_{u})\underbrace {cosx\, dx}_{dv}=[du=-dx\; ,\; v=sinx\; ]=uv-\int v\, du=\\\\=(3-x)\, sinx-\int sinx(-dx)=(3-x)\, sinx-cosx+C\; ;\\\\3)\; \; \int \frac{\sqrt3\, dx}{9x^2-3}=\int \frac{\sqrt3\, dx}{(3x)^2-(\sqrt3)^2}=\sqrt3\cdot \frac{1}{2\sqrt3}\cdot \frac{1}{3}\cdot ln\Big |\frac{3x-\sqrt3}{3x+\sqrt3}\Big |+C=\\\\=\frac{1}{6}\cdot ln\Big |\frac{3x-\sqrt3}{3x+\sqrt3}\Big |+C\; ;$

$4)\; \; \int e^{1-4x^2}\cdot x\, dx=[\; u=1-4x^2\; ,\; du=-8x\, dx\; ]=\\\\=-\frac{1}{8}\int e^{u}\, du=-\frac{1}{8}\cdot e^{u}+C=-\frac{1}{8}\cdot e^{1-4x^2}+C\; .\\\\\\P.S.\; \; \int f(x)\, dx=F(x)+C\; \; \Rightarrow \; \; \int f(kx+b)\, dx=\frac{1}{k}\cdot F(kx+b)+C\; .$