Question

Решить интегралы, 5 и 6

Answer 1

$1)\; \; \int x\cdot ctg^2x\cdot dx=\int x\cdot (\frac{1}{sin^2x}-1)\, dx=\int \frac{x\, dx}{sin^2x}-\int x\, dx=(\star )\\\\\\\int \frac{x\, dx}{sin^2x}=[\; u=x\; ,\; du=dx\; ,\; dv=\frac{dx}{sin^2x}\; ,\; v=-ctgx\; ]=\\\\=uv-\int v\cdot du=-x\cdot ctgx+\int ctgx\, dx=-x\cdot ctgx+\int \frac{cosx\, dx}{sinx}=\\\\=-x\cdot ctgx+\int \frac{s(sinx)}{sinx}=-x\cdot ctgx+ln|sinx|+C_1\; ;\\\\\\(\star )=-x\cdot ctgx+ln|sinx|-\frac{x^2}{2}+C\; ;$

$2)\; \; \int \frac{x^3+2x}{x^4+4x^2}\, dx=\int \frac{x(x^2+2)}{x^2(x^2+4)}\, dx=\int \frac{x^2+2}{x\, (x^2+4)}\, dx=(\star )\\\\\\\frac{x^2+2}{x\, (x^2+4)}=\frac{A}{x}+\frac{Bx+C}{x^2+4}=\frac{A(x^2+4)+x\cdot (Bx+C)}{x\, (x^2+4)}\\\\x^2+2=Ax^2+4A+Bx^2+Cx\\\\x^2\; |\; 1=A+B\\x\; \; |\; 0=C\\x^\circ \; |\; 2=4A\; \; ,\; \; A=\frac{1}{2}\; \; ,\; \; B=1-A=\frac{1}{2}$

$(\star )\; =\int \frac{\frac{1}{2}}{x}\, dx+\frac{1}{2}\int \frac{x\, dx}{x^2+4}=\frac{1}{2}\cdot ln|x|+\frac{1}{4}\int \frac{2x\, dx}{x^2+4}=\\\\=\frac{1}{2}\cdot ln|x|+\frac{1}{4}\int \frac{d(x^2+4)}{x^2+4}=\frac{1}{2}\cdot ln|x|+\frac{1}{4}\cdot ln|x^2+4|+C$

$3)\; \; \int\limits^{e^4}_1 \, \frac{\sqrt[3]{4+lnx}}{x}\, dx=[\; d(4+lnx)=(4+lnx)'\, dx=\frac{dx}{x}\; ]=\\\\=\int\limits^{e^4}_1 \, \sqrt[3]{4+lnx}\, \cdot d(4+lnx)=\frac{(4+lnx)^{\frac{4}{3}}}{\frac{4}{3}}\Big |_1^{e^4}=\\\\=\frac{3}{4}\Big ((4+lne^4)^{4/3}-(4+ln1)^{4/3}\Big )=\frac{3}{4}((4+4)^{4/3}-(4+0)^{4/3})=\\\\=\frac{3}{4}(\sqrt[3]{8^4}-\sqrt[3]{4})=\frac{3}{4}(2^4-\sqrt[3]4)\approx \frac{3}{4}(16-1,59)=10,8075$