Question

Решите кто понимает очень хочу понять как решается подобное.

Answer 1

$1)\; \; \int \frac{(6x-5)\, dx}{2\sqrt{3x^2-5x+6}}=\Big [\; 3x^2-5x+6=3\, (x^2-\frac{5}{3}x+2)=3\, \Big ((x-\frac{5}{6})^2-\frac{25}{36}+2\Big )=\\\\=3\, \Big ((x-\frac{5}{6})^2- \frac{47}{36}\Big )\; \Big]=\frac{1}{2\sqrt3}\int \frac{(6x-5)\, dx}{\sqrt{(x-\frac{5}{6})^2-\frac{47}{36}}}=\Big[\; t=x-\frac{5}{6}\; ,\; dt=dx,$

$x=t+\frac{5}{6}\; \Big]=\frac{1}{2\sqrt3}\int \frac{6t\, dt}{\sqrt{t^2-\frac{47}{36}}}=\frac{3}{2\sqrt3}\int \frac{2\, t\, dt}{\sqrt{t^2-\frac{47}{36}}}=\Big [\; z=t^2-\frac{47}{36}\; ,\; dz=2t\, dt\; \Big]=\\\\=\frac{\sqrt3}{2}\int \frac{dz}{\sqrt{z}}=\frac{\sqrt3}{2}\cdot 2\sqrt{z}+C=\sqrt3\cdot \sqrt{x^2-\frac{5}{3}x+2}+C\; ;$

$2)\; \; \int \frac{\sqrt{1-x^2}}{x^2}\, dx=\Big[\; x=sint\; ,\; dx=cost\, dt\; ,\; t=arcsinx\; ,\\\\\sqrt{1-x^2}=\sqrt{1-sin^2t}=\sqrt{cos^2t}=cost\; \Big]=\int \frac{cost}{sin^2t}\cdot cost\, dt=\\\\=\int \frac{cos^2t}{sin^2t}\, dt=\int \frac{1-sin^2t}{sin^2t}\, dt=\int \frac{dt}{sin^2t}-\int dt=-ctgt-t+C=\\\\=-ctg(arcsinx)-arcsinx+C=\frac{\sqrt{1-x^2}}{x}-arcsinx+C$

$3)\; \; \int \frac{cos^3x}{sin^2x+sinx}\, dx=\int \frac{cos^2x\cdot cosx\, dx}{sin^2x(sinx+1)}=\int \frac{(1-sin^2x)\cdot cosx\, dx}{sinx(sinx+1)}=\\\\\Big[\; t=sinx\; ,\; dt=cosx\, dx\; \Big]=\int \frac{(1-t^2)\cdot dt}{t(t+1)}=\int \frac{(1-t)(1+t)\cdot dt}{t\cdot (t+1)}=\int \frac{1-t}{t}\, dt=\\\\=\int (\frac{1}{t}-1)\, dt=\int \frac{dt}{t}-\int dt=ln|t|-t+C=ln|sinx|-sinx+C\; ;$

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