Математика, найти первообразные

Question 1

Question 2

Решите задачу:

$\displaystyle 4)\int\frac{dx}{arctgx(x^2+1)}=\frac{d(arctgx)}{arctgx}=ln|arctgx|+C\\\\5)\int\frac{dx}{x^2-6x+6}=\int\frac{d(x-3)}{(x-3)^2-3}=\frac{1}{2\sqrt3}ln|\frac{x-3-\sqrt3}{x-3+\sqrt3}|+C$
$\displaystyle6)\int\frac{(2x-3)dx}{x^2-2x-3}=\int\frac{(2x-2-1)dx}{x^2-2x-3}=\int\frac{d(x^2-2x-3)}{x^2-2x-3}-\\-\int\frac{dx}{(x-3)(x+1)}=\int\frac{d(x^2-2x-3)}{x^2-2x-3}-\frac{1}{4}\int\frac{dx}{x-3}+\frac{1}{4}\int\frac{dx}{x+1}=\\=ln|x^2-2x-3|-\frac{1}{4}(ln|x-3|-ln|x+1 |)=\\\\=ln|x^2-2x-3|-\frac{1}{4}ln|\frac{x-3}{x+1} |+C\\\\\\\frac{1}{(x-3)(x+1)}=\frac{A}{x-3}+\frac{B}{x+1}=\frac{1}{4(x-3)}-\frac{1}{4(x+1)}\\1=A(x+1)+B(x-3)\\x^0|1=A-3B\\x^1|0=A+B=\ \textgreater \ A=-B\\1=-B-3B\\B=-\frac{1}{4}\\A=\frac{1}{4}$
$\displaystyle7)\int(4-5x)e^xdx\\u=4-5x=\ \textgreater \ du=-5dx\\dv=e^xdx=\ \textgreater \ v=e^x\\\int(4-5x)e^xdx=(4-5x)e^x+5\int e^xdx=(4-5x)e^x+5e^x+C=\\=e^x(9-5x)+C$

Question 3

Решите задачу:

$4)\; \; \int \frac{dx}{arctgx\cdot (x^2+1)} =\int \frac{\frac{dx}{x^2+1}}{arctgx} =\int \frac{d(arctgx)}{arctgx}=\Big [\int \frac{dt}{t}\Big ]=ln|arctgx|+C\\\\5)\; \; \int \frac{dx}{x^2-6x+6} =\int \frac{dx}{(x-3)^2-3} =\int \frac{d(x-3)}{(x-3)^2-3} =\Big [\, \int \frac{dt}{t^2-a^2}\Big ]= \\\\=\frac{1}{2\sqrt{3}}\cdot ln\Big |\frac{x-3-\sqrt{3}}{x-3+\sqrt{3}} \Big |+C$

$6)\; \; \int \frac{2x-3}{x^2-2x-3} \, dx=\int \frac{2x-3}{(x-1)^2-4}\, dx=\Big [\, t=x-1,\; x=t+1,\; dx=dt\Big ]=\\\\= \int \frac{2t-1}{t^2-4}=\int \frac{2t\, dt}{t^2-4}-\int \frac{dt}{t^2-4} =\int \frac{d(t^2-4)}{t^2-4}-\frac{1}{2\cdot 2} \cdot ln\Big | \frac{t-2}{t+2}\Big |=[\int \frac{du}{u}]=\\\\= ln|t^2-4|- \frac{1}{4}\cdot ln\Big |\frac{t-2}{t+2}\Big |+C= ln|x^2-2x-3|- \frac{1}{4}\, ln\Big |\frac{x-3}{x+1} \Big |+C$

$7)\; \; \int (4-5x)\; e^{x}dx=[\, u=4-5x,\; du=-5\, dx,\; dv=e^{x}\, dx,\; v=e^{x}]=\\\\=uv-\int v\, du=(4-5x)e^{x}+5\cdot \int e^{x}\, dx=\\\\=(4-5x)\, e^{x}+5e^{x}+C=e^{x}\, (4-5x+5)+C=(9-5x)\, e^{x}+C$