Question

По интеграллам int (x+5)^3dx int x(x^2-1)^3dx int(x^2+5)^7 2xdx int xdx/x^2 11 int dx/(x-1)^4 int√(1+2x) dx заранее огромное !

Answer 1

$\int(x+5)^3dx=\left(\begin{array}{c}u=x+5\\du=dx\end{array}\right)=\int u^3du=\frac14u^4+C=\frac14(x+5)+C\\\\\int x(x^2-1)^3dx=\left(\begin{array}{c}u=x^2-1\\du=2x\;dx\end{array}\right)=\frac12\int u^3du=\frac18u^4+C=\\=\frac18(x^2-1)^4+C\\\\\int2x\cdot(x^2+5)^7\;dx=\left(\begin{array}{c}u=x^2+5\\du=2x\;dx\end{array}\right)=\int u^7du=\frac18u^8+C=\\=\frac18(x^2+5)^8+C\\\\\int\frac{x\;dx}{x^2+1}=\left(\begin{array}{c}u=x^2\\du=2x\;dx\end{array}\right)=\frac12\int\frac1u\;du=\frac12\ln u+C=\frac12\ln(x^2+1)+C$

$\int\frac{dx}{(x-1)^4}=\left(\begin{array}{c}u=x-1\\du=dx\end{array}\right)=\int\frac{du}{u^4}=-\frac1{u^3}+C=-\frac1{(3(x-1)^3)}+C$

$\int\sqrt{1+2x}dx=\left(\begin{array}{c}u=1+2x\\du=2dx\end{array}\right)=\frac12\int\sqrt u\;du=\frac13u^{\frac32}+C=\\=\frac13(1+2x)^{\frac32}+C$