Question

Нужно решение 7 любых примеров (неопределенный интеграл)

Answer 1

$\displaystyle \int \frac{dx}{x(lnx+5)}=\int\frac{d(lnx+5)}{lnx+5}=ln|lnx+5|+C$

$\displaystyle \int cos^3xsin^3xdx=\frac{1}{16}\int sin^32x\ d(2x)=\\=-\frac{1}{16}\int( 1-cos^22x)d(cos2x)=-\frac{1}{16}(cos2x-\frac{cos^32x}{3})+C$

$\displaystyle \int \frac{(x-3)dx}{4x^2+4x-9}=\frac{1}{8}\int\frac{8x+4-28}{4x^2+4x-9}dx=\frac{1}{8}\int\frac{d(4x^2+4x-9)}{4x^2+4x-9}-\\-\frac{7}{4}\int\frac{d(2x+1)}{(2x+1)^2-10}=\frac{1}{8}ln|4x^2+4x-9|-\\-\frac{7}{8\sqrt{10}}ln|\frac{2x-1-\sqrt{10}}{2x-1+\sqrt{10}}|+C\\\\\\(4x^2+4x-9)'=8x+4$

$\displaystyle \int \sqrt[3]{x}lnxdx=\frac{3\sqrt[3]{x^4}lnx}{4}-\frac{3}{4}\int\sqrt[3]{x}dx=\frac{3\sqrt[3]{x^4}lnx}{4}-\frac{9\sqrt[3]{x^4}}{16}+C\\u=lnx;du=\frac{dx}{x}\\dv=\sqrt[3]{x}dx;v=\frac{3\sqrt[3]{x^4}}{4}$

$\displaystyle \int \frac{dx}{\sqrt{1-2x}+3}=-\int\frac{(t+3-3)dt}{t+3}=-\int(1-\frac{3}{t+3})dt=\\=-t+3ln|t+3|+C=-\sqrt{1-2x}+3ln|\sqrt{1-2x}+3|+C\\1-2x=t^2\\x=\frac{1-t^2}{2}\\dx=-tdt$

$\displaystyle \int \frac{x^5dx}{x^3-1}=\int\frac{((x^3-1)x^2+x^2)dx}{x^3-1}=\int(x^2+\frac{x^2}{x^3+1})dx=\\=\int x^2dx+\frac{1}{3}\int\frac{d(x^3+1)}{x^3+1}=\frac{x^3}{3}+\frac{ln|x^3+1|}{3}+C$

$\displaystyle \int (x-3)sinxdx=(3-x)cosx+\int cosxdx=(3-x)cosx+sinx+C\\u=(x-3);du=dx\\dv=sinxdx;v=-cosx$

$\displaystyle \int sin^5xdx=-\int (1-cos^2x)^2\ d(cosx)=\\=-\int(1-2cos^2x+cos^4x)\ d(cosx)=-cosx+\frac{2cos^3x}{3}-\frac{cos^5x}{5}+C$

$\displaystyle \int \frac{arcsinx}{\sqrt{1-x^2}}dx=\int arcsinx\ d(arcsinx)=\frac{arcsin^2x}{2}+C$

$\int x*2^{x-1}dx=\frac{x*2^{x-1}}{ln2}-\frac{1}{ln2}\int 2^{x-1}dx=\frac{x*2^{x-1}}{ln2}-\frac{2^{x-1}}{ln^22}+C\\u=x;du=dx\\dv=2^{x-1}dx;v=\frac{2^{x-1}}{ln2}$