2ч.:з Все те же интегралы. (45б)

Решите задачу:

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