Question

Помогите пожалуйста решить

Answer 1

$1)\; \; \int 3x^2(1-x^3)^8\, dx=[\; t=1-x^3\; ,\; dt=-3x^2\, dx\; ]=-\int t^8\, dt=\\\\=-\frac{t^9}{9}+C=-\frac{(1-x^3)^9}{9}+C\\\\\\2)\; \; \int \frac{3-2x}{5x^2+7}\, dx=\int \frac{3\, dx}{5x^2+7}-\frac{1}{5}\int \frac{10x\, dx}{5x^2+7}=\frac{3}{\sqrt5}\int \frac{d(\sqrt5x)}{(\sqrt5x)^2+(\sqrt7)^2}-\frac{1}{\sqrt5}\int \frac{d(5x^2+7)}{5x^2+7}=\\\\=\frac{3}{\sqrt5}\cdot \frac{1}{\sqrt7}\cdot arctg\frac{\sqrt5x}{\sqrt7}-\frac{1}{\sqrt5}\cdot ln|5x^2+7|+C$

$3)\; \; \int \frac{dx}{4+9x^2}=\frac{1}{3}\int\frac{3\, dx}{2^2+(3x)^2}=\frac{1}{3}\cdot \frac{1}{2}\cdot arctg\frac{3x}{2}+C=\frac{1}{6}\cdot arctg\frac{3x}{2}+C$

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