这道不定积分怎么做

配方就可以了

integral 1/sqrt(2 t^2+2 t+1) dt = (sinh^(-1)(2 t+1))/sqrt(2)+constant

会了吗？

Take the integral:
integral 1/sqrt(2 t^2+2 t+1) dt
For the integrand 1/sqrt(2 t^2+2 t+1), complete the square:
= integral 1/sqrt((sqrt(2) t+1/sqrt(2))^2+1/2) dt
For the integrand 1/sqrt((sqrt(2) t+1/sqrt(2))^2+1/2), substitute u = sqrt(2) t+1/sqrt(2) and du = sqrt(2) dt:
= 1/sqrt(2) integral 1/sqrt(u^2+1/2) du
For the integrand 1/sqrt(u^2+1/2), substitute u = (tan(s))/sqrt(2) and du = (sec^2(s))/sqrt(2) ds. Then sqrt(u^2+1/2) = sqrt((tan^2(s))/2+1/2) = (sec(s))/sqrt(2) and s = tan^(-1)(sqrt(2) u):
= 1/2 integral sqrt(2) sec(s) ds
Factor out constants:
= 1/sqrt(2) integral sec(s) ds
The integral of sec(s) is log(tan(s)+sec(s)):
= (log(tan(s)+sec(s)))/sqrt(2)+constant
Substitute back for s = tan^(-1)(sqrt(2) u):
= (log(tan(tan^(-1)(sqrt(2) u))+sec(tan^(-1)(sqrt(2) u))))/sqrt(2)+constant
Simplify using sec(tan^(-1)(z)) = sqrt(z^2+1) and tan(tan^(-1)(z)) = z:
= (log(sqrt(2 u^2+1)+sqrt(2) u))/sqrt(2)+constant
Substitute back for u = sqrt(2) t+1/sqrt(2):
= (

划到这了我还是不会