59. ∫ dx / √ ( x + 1 + √(x + 1)³)
√1–x=t
1–x=t2
x=1–t2
2–x=1+t2
dx=–2tdt
= ∫ (–2tdt)/((1+t2)·t)=–2 ∫ dt/(1+t2)=–2arctgt+C=
=–2arctg√x–1 + C
59
√x+1=t
√(x+1)3=t3
x+1=t2
x=t2–1
dx=2tdt
= ∫ 2tdt/(t+t3)=2 ∫ dt/(1+t2)=2arctg(t) + C= 2 arctg√x+1 + C