`int{log(logx)+(1)/((logx)^(2))}dx=x {f (x)-g(x)}+c`, then
`f(x)=log(logx),g (x)=(1)/(logx)`
`f(x)=logx,g(x)=(1)/(logx)`
`f(x)=(1)/(logx),(x)=log(logx)`
`f(x)=(1)/(xlogx),g(x)=(1)/(logx)`
answer: a
solution: we have , `i=int{log(logx)+(1)/((logx)^(2))}dx`   
 `rarr i= inte^(t)(logt+(1)/(t^(2)))dt`,where t = log x   
 `rarri=inte^(t)(logt+(1)/(t))dt+inte^(t)(-(1)/(t)+(1)/(t^(2)))dt`   
 `rarri=e^(t)logt +e^(t)(-(1)/(t))+c`   
 `rarri=x(log(logx)-(1)/(logx))+c`   
 `:. f(x) =logx and g(x) = (1)/(logx)`
given = let, logx = z⇒ so,where c is the integrating constant. let i = `(1)/(sinx*(3 + 2cosx))*dx` = `int sinx/(sin^2x*(3 + 2cosx))*dx` = `int sinx/((1 - cos^2x)(3 + 2cosx))*dx` = `int sinx/((1 - cosx)(1 + cosx)(3 + 2cosx))*dx` put cos x = t∴ – sinx.dx = dt∴ sinx.dx = – dt ∴ i = `int (1)/((1 - t)(1 + t)(3 + 2t))*(-dt)` = `int (-1)/((1 - t)(1 + t)(3 + 2t))*dt` let `(-1)/((1 - t)(1 + t)(3 + 2t)) = "a"/(1 - t) + "b"/(1 + t) + "c"/(3 + 2t)` ∴ – 1 = a(1 + t)(3 + 2t) + b(1 - t)(3 + 2t) + c(1 - t)(1 + t)put 1 – t = 0, i.e. t = 1, we get– 1 = a(2)(5) + b(0)(5) + c(0)(2) ∴ – 1 = 10a ∴ a = `(-1)/(10)`put 1 + t = 0, i.e. t = – 1, we get– 1 = a(0)(1) + b(2)(1) + c(2)(0) ∴ – 1 = 2b ∴ b = `-(1)/(2)` put 3 + 2t = 0, i.e. t = `-(3)/(2)`, we get – 1 = `"a"(-1/2)(0) + "b"(5/2)(0) + "c"(5/2)(-1/2)` ∴ – 1 = `-(5)/(4)"c"` ∴ c = `(4)/(5)` ∴ `(-1)/((1 - t)(1 + t)(3 + 2t)) = (((-1)/(10)))/(1 - t) + ((-1/2))/(1 + t) + ((4/5))/(3 + 2t)` ∴ i = `int [(((-1)/10))/(1 - t) + ((-1/2))/(1 + t) + ((4/5))/(3 + 2t)]*dt` = `-(1)/(10) int 1/(1 - t)*dt - (1)/(2) int 1/(1 + t)*dt + (4)/(5) int 1/(3 + 2t)*dt` = `-(1)/(10) (log|1 - t|)/(-1) - (1)/(2) log | 1 + t| + 4/5 (log|3 + 2t|)/(2) + c` = `(1)/(10)log|1 - cosx| - (1)/(2)log|1 + cosx| + (2)/(5)log|3 + 2cos| + c`. formula :- therefore ,put -2x dx = dt doing integration yields-then finally f(x)=-x cos x + sin xto calculate derivative of the above function f(x) we have to apply formula of derivation of products of two functions-f(x) =-x cos(x)+sin(x)f’(x) =-[x(-sin x)+cos x.1]+cos x ;{by formula d/dx (f.g)=f.g’+g.f’}f’(x) =-(-x sin x) - cos x + cos xf’(x) = x.sin(x) given : put ln x = t 1) p(x) = 1/x2 + 1 if x ≤ 3 it is clear that x= 3 is included, so you have to use this function.2) loo k at table d, and compare the given information with the data written inside the table. c = input()  

t = len(c)  

temp = ""  

for i in range(0,len(c)):  

    if t<3:  

        break  

    else:  

        if i in (0,1,t-2,t-1):  

            temp = temp+c[i]  

        else:  

            continue  

print(temp) if , passing through (1, 2), satisfies the differential equation , then log 80 can be written as; log (8 ✕ 10) log (4 ✕ 2 ✕ 10) here we can use the formula log (a ✕ b) = log a + log b log 4 + log 2 + log 10  (1) log10 10 = 1 log 4 = 0.6020 log 2 = ½ log 4 = ½ ✕ 0.6020 = 0.3010 substituting the values in equation (1) we get the answer 1.9030 f = ma(t) = kte^(-at) => a(t) = k/m * te^(-at)

v(t) = ∫a(t) dt = k/(ma^2) * (-ate^(-at) - e^(-at) + c)
= -k/(ma^2) * e^(-at)*(at + 1 + ce^(at))