Question

Найти интегралы.
Заранее спасибо

Answer 1

$1)\; \; \int \frac{dx}{3x^2+4}=\int \frac{dx}{(\sqrt3x)^2+2^2}=\frac{1}{2\sqrt3}\cdot arctg\frac{\sqrt3x}{2}+C\\\\2)\; \; \int \frac{e^{3x}\, dx}{e^{3x}+4}=[\; u=e^{3x}+4\; ,\; du=3e^{3x}\, dx\; ]=\frac{1}{3}\int \frac{du}{u}=\frac{1}{3}\cdot ln|u|+C=\\\\=\frac{1}{3}\cdot ln|e^{3x}+4|+C\\\\3)\; \; \int \frac{4-6x^3}{x^2-x+5}\, dx=\int \Big (-6x-6+\frac{24x+34}{x^2-x+5}\Big )\, dx=\\\\=\int \Big (-6x-6+\frac{24x+34}{(x-\frac{1}{2})^2+\frac{19}{4}}\Big )\, dx=-6\int x\, dx-6\int \, dx+\int \frac{24(x-\frac{1}{2})+46}{(x-\frac{1}{2})^2+\frac{19}{4}}\, dx=$

$=[\; t=x-\frac{1}{2}\; ,\; x=t+\frac{1}{2}\; ,\; dt=dx\; ]=\\\\=-6\cdot \frac{x^2}{2}-6x+12\int \frac{2t\, dt}{t^2+\frac{19}{4}}+46\int \frac{dt}{t^2+\frac{19}{4}}=-3x^2-6x+12\cdot ln\Big |t^2+\frac{19}{4}\Big |+\\\\+46\cdot \frac{2}{\sqrt{19}}\cdot arctg\frac{2t}{\sqrt{19}}+C=-3x^2-6x+12\cdot ln\Big |x^2-x+5\Big |+\\\\+\frac{92}{\sqrt{19}}\cdot arctg\frac{2x-1}{\sqrt{19}}+C$

$4)\; \; \int \limits _0^{1/4}arctg4x\, dx=[\; u=arctg4x\; ,\; dv=dx\; ,\; v=x\; ,\; du=\frac{4\, dx}{1+16x^2}\; ]=\\\\=x\cdot arctg4x\Big |_0^{1/4}-\int\limits^{1/4}_0\, \frac{4x\, dx}{1+16x^2}=\frac{1}{4}\cdot \frac{\pi}{4}-\frac{1}{8}\int \limits _0^{1/4}\, \frac{d(1+16x^2)}{1+16x^2}=\\\\=\frac{\pi}{16}-\frac{1}{8}\cdot ln|1+16x^2|\Big |_0^{1/4}=\frac{\pi}{16}-\frac{1}{8}\cdot ln2$