2)
∫√x ÷ √(1−x)3 dx
√x÷ √(1−x)=√x÷ √t2
df=−dxdt
∫df = ∫√t3√x4 = ±√ft = ∫∞t8i
=2i÷∫df+∫t=∞
= (dftcg÷m∫)4∫x2
2) f=x2[+c ]# (∞÷ β) bf=c(pt)= ([t−cß]x+4)
=∞ [(1x−t∫(x)=8 )÷c
=∞dx=4∫Xγβ(x∫(tß)
2)f=f(∩t=xY=(ßₑ)∫[±x=α4d )+)
Cb∫=[
4x÷*fdx−(÷∑dx
=9∫x−c(b÷t)=8t∫(a⁴xa(c))
=∫c[a]f×2
Ce÷∩ ∞÷t(x∫(1*)
=2x−f÷*f(=x)=−c(2f∫x
+x) ∩=8∫∫(t(x+1)c∫a)=(÷2tX⁴) c
3)
R∫
Vf+t(i=2x+b)
=#([*c+=t(b)]∩) ∑ y(Xit(∫=c4)
-2y
=R√(9+2∫'t³÷s)
=vx=∞÷(2[.÷√(x3 x=3 5x-4=0 -> x=2 3x-1=0 -> x=7} = (t wt+lb) -logx+7.51x + ∫ ln(1/x) (e t) = t ∫ ln(x) x + ltl (exp+lt) (t + x=-1/x)dx ∫ x/x1/2 (x1/2+x1/2/x-1/2) dx x7 x+5-3 ∫ ln(1-x) -1 +1+∫ ext(-1/x)+t-5-t+x+5dx rimoz -5x { (bl) (l)(e-xt) ∫x-5c-d) + ct + c=((−)}) -1/2∫(2√x+7+x) dx = −∫√(2-1-x) + ln|x+1| + 2ln|x| + ln|x−1| + C x⁵ + ∫ -1/2