Question

1 курс математика ИНТЕГРАЛЫ

Answer 1

$1)\; \; \int x^2\, (7-2x^3)\, dx=\Big [\; u=7-2x^3\; ,\; du=-6x^2\, dx\; \Big ]=\\\\=-\frac{1}{6}\int (7-2x^3)\cdot (-6x^2)\, dx=-\frac{1}{6}\, \int u\cdot du=-\frac{1}{6}\cdot \frac{u^2}{2}+C=-\frac{1}{12}\cdot (7-2x^3)^2+C\; ;\\\\\\2)\; \; \int sin4x\cdot 2^{cos^22x+1}\cdot dx=\Big [\; u=cos^22x+1\; ,\; du=2\, cos2x\cdot (-sin2x)\cdot 2\cdot dx=\\\\=-2\cdot sin4x\cdot dx\; \Big ]=-\frac{1}{2}\int 2^{u}\cdot du=-\frac{1}{2}\cdot \frac{2^{u}}{ln2}+C=-\frac{1}{2\cdot ln2}\cdot 2^{cos^22x+1}+C\; ;$

$3)\; \; \int \frac{x\cdot 2^{x^2-1}}{\sqrt{4^{x^2}+1}}=\int \frac{2^{x^2}\cdot x\, dx}{2\cdot \sqrt{(2^{x^2})^2+1}}=\Big [\; u=2^{x^2}\; ,\; du=2^{x^2}\cdot ln2\cdot 2x\cdot dx\; \Big ]=\\\\=\frac{1}{2\, ln2}\cdot \int \frac{du}{2\cdot \sqrt{u^2+1}}=\frac{1}{4\, ln2}\int \frac{du}{\sqrt{u^2+1}}=\frac{1}{4\, ln2}\cdot ln|\, u+\sqrt{u^2+1}}\, |+C=\\\\=\frac{1}{ln16}\cdot ln\Big |\, 2^{x^2}+\sqrt{4^{x^2}+1}\, \Big |+C\; .$