Russian College Test Integral Problem

[@renesperb]

Another way: set x/√7 = y = sinh[s] ,i,e, dy =cosh[s] ds. Since √(1+sinh[s] )^2 = cosh[s] the integral becomes just s .Going back to the variable x we get the answer arsinh[x/√7 ] .

[@holyshit922]

They would probably used substitution sqrt(x^2+7) = u - x and guy who came with this substitution lived among them

[@paulvalck349]

Int(1/sqrt(x²+7).dx) let u = x + sqrt(x²+7) then du = (1+x/sqrt(x²+7)).dx du = ((sqrt(x²+7)+x)/sqrt(x²+7)).dx du = u/(sqrt(x²+7)).dx du/u = 1/sqrt(x²+7).dx The given integral is equal to Int(du/u) = ln|u| + c Int(1/sqrt(x²+7).dx) = ln|x + sqrt(x²+7)| + c This is equal to the answer in the video as ln|(x+sqrt(x²+7))/sqrt(7)| + c = ln|x+sqrt(x²+7)| - ln(sqrt7)) + c = ln|x+sqrt(x²+7)| + c' c' = c - ln(sqrt(7))

[@dan-florinchereches4892]

Nice solution. I was just thinking of going standard and just using Euler substitution x+t=sqrt(x^2+7) but i guess your variant is the one taking -x+t=sqrt(x^2+7) your calculations ended up really nice