Question

find the inverse Laplace transformation of 1/(s+1)(s^2+1)​

Answer 1

Answer:

your answer is in this page

Step-by-step explanation:

Please mark me as the branlist

Answer 2

Find the inverse transform of the given function.

Explanation:

• For a function existing in the s - domain there exist its corresponding function in the time domain.
• This corresponding time domain function can be obtained from its s - domain function by inverse Laplace transformation and is given by, $f(t)=\L ^{-1} F(s)=\frac{1}{2\pi i} \lim_{T\to \infty} \int\ e^{st}F(s)ds \, dx$  
• Given function in s domain, $F(s)=\frac{1}{(s+1)(s^2+1)}$    
• We know that for given below s-domain functions ,                                         $F(s)=\frac{1}{s+1}\ \ \ ->f(t)=e^{-1} , \\ F(s)=\frac{1}{s^2+1}\ \ \ ->f(t)=sint \\ F(s)=\frac{s}{s^2+1} \ \ \ ->f(t)=cost$    
• from the above given functions we get the inverse transformation as,                     $f(t)=\frac{1}{2} (sint-cost+e^{-t})$      ---->ANSWER