Интегралы .Номера 3.20, 3.21, 3,22, 3,23,3.4

Question 1

Question 2

3.20. $\int {\frac{ln(x)}{\sqrt{x}}}\,dx=\left[\begin{array}{cc}u=ln(x)&dv=\frac{dx}{\sqrt{x}}\\du=\frac{dx}{x}&v=2 \sqrt{x}\end{array}\right]=2\sqrt{x}*ln(x)-\int {\frac{2}{\sqrt{x}}}\,dx=$ $2\sqrt{x}*ln(x)-4\sqrt{x}+C$
3.21. $\int {(2x-3)sin\frac{x}{2}\,dx=\left[\begin{array}{cc}u=2x-3&dv=sin\frac{x}{2}dx\\du=2dx&v=-2cos\frac{x}{2}\end{array}\right]=$ $2(3-2x)cos\frac{x}{2}+4\int\cos\frac{x}{2}\, dx=2(3-2x)cos\frac{x}{2}+8sin\frac{x}{2}+C=$ $2(3-2xcos\frac{x}{2}+4sin\frac{x}{2})+C$
3.22. $\int{ln^2x}\,dx=\left[\begin{array}{cc}u=ln^2x&dv=dx\\du=\frac{2ln(x)dx}{x}&v=x\end{array}\right]=x*ln^2x-2\int{ln(x)}\,dx=$ $\left[\begin{array}{cc}u=ln(x)&dv=dx\\du=\frac{dx}{x}&v=x\end{array}\right]=x*ln^2x-2(x*ln(x)-\int dx)=x*ln^2x-2(x*ln(x)-x)+C=x(ln^2x-2ln(x)+2)+C$
3.23. $\int{\frac{ln(x)}{(x+1)^2}}\,dx=\left[\begin{array}{cc}u=ln(x)&dv=\frac{dx}{(x+1)^2} \\du=\frac{dx}{x}&v=-\frac{1}{x+1}\end{array}\right]=-\frac{ln(x)}{x+1}+\int{\frac{dx}{x(x+1)}=-\frac{ln(x)}{x+1}+\int{\frac{dx}{x}-\int{\frac{dx}{x+1}$ $=-\frac{ln(x)}{x+1}+ln(x)-ln(x+1)+C$
3.24. $\int{cos(ln(x))}dx=\left[\begin{array}{cc}u=cos(ln(x))&dv=dx\\du=-\frac{sin(ln(x))}{x}dx&v=x}\end{array}\right]=$ $x*cos(ln(x))+\int{sin(ln(x))}dx=\left[\begin{array}{cc}u=sin(ln(x))&dv=dx\\du=\frac{cos(ln(x))}{x}dx&v=x}\end{array}\right]=$ $=x*cos(ln(x))+x*sin(ln(x))-\int cos(ln(x))dx\\\int cos(ln(x))dx=\frac{x(sin(ln(x))+cos(ln(x)))}{2}+C$

Question 3

Решите задачу:

$3.20) \int\limits { \frac{lnx}{ \sqrt{x} } } \, dx =|u=lnx||dv= \frac{1}{ \sqrt{x} } dx||du= \frac{dx}{x}||v=2 \sqrt{x} |=2 \sqrt{x} *lnx \\ - 2\int\limits{ x^{- \frac{1}{2} } } \, dx=2 \sqrt{x} *lnx-4 \sqrt{x} +C \\ 3.21) \int\limits {(2x-3)sin\frac{x}{2} } \, dx =|u=2x-3||dv=sin \frac{x}{2}dx||du=2| \\ |v=-2cos \frac{x}{2} |=(2x-3)*(-2cos \frac{x}{2})+4 \int\limits {cos \frac{x}{2} } \, dx = \\ =(2x-3)*(-2cos \frac{x}{2})+8sin \frac{x}{2}+C$