number of integers satisfying the inequality `log\_((x+3))(x^2-x) lt 1` is
answer: `1.00`
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`. multiply 2 in both lhs and rhs,integrating on both the sides, write a pair of intergers whose sum gives hi,

sorry you are having issues with this. this is a known bug that was fixed in 2.24. the index of the loop is of an unsigned type in the generated c++ in 2.23. and since there are no negative numbers in unsigned types -1 becomes max integer.

issue where we found this: [https://github.com/stan-dev…](https://github.com/stan-dev/stanc3/issues/536) given : put ln x = t 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) 2, 0 ( any two integers with one of them as 0). 65 ÷ ( – 13) = **-5** hello, red.since the triangle is a right triangle, you can use the pythagorean theorem.let x be the length of the unknown side.then,92 + x2 = (9√2)281 + x2 = 81·2 = 162x2 = 162 - 81x2 = 81take the square root of both sides.x = ±9since the length cannot be a negative number, x = 9 m