Question

2ч.:з Все те же интегралы.
(45б)

Answer 1

$int frac{3dx}{(x+1)^4}=3cdot int (x+1)^{-4}dx=3cdot frac{(x+1)^{-3}}{-3}+C=-frac{1}{(x+1)^3}+C\\int 2^{x^4}cdot x^3dx=[t=x^4; ,dt=4x^3dx]=frac{1}{4}int 2^{t}dt=frac{1}{4}cdot frac{2^{t}}{ln2}+C=frac{2^{x^4}}{4ln2}+C\\int frac{e^{ctgx}}{sin^2x}dx==[t=ctgx,; dt=-frac{dx}{sin^2x}]=-int e^{t}dt=-e^{t}+C=\\=-e^{ctgx}+C$