To finish a Laplace problem you must go back from $F(s)$ to $f(t)$. There is no new integral to learn: you rewrite $F$ as a sum of table entries, usually with partial fractions, and read the table backwards.
Key idea
$\mathcal{L}^{-1}$ is linear. Split $F$ into partial fractions; linear factors give exponentials, repeated factors give $t^{n}e^{at}$, and irreducible quadratics are handled by completing the square and matching $$\frac{s - a}{(s - a)^{2} + b^{2}} \to e^{at}\cos bt,$$ $$\frac{b}{(s - a)^{2} + b^{2}} \to e^{at}\sin bt.$$
Why it works
Two continuous functions with the same transform are equal (Lerch's theorem), so reading the table backwards gives the one continuous answer. Partial fractions work because every proper rational function is a sum of simple pieces, and each piece is a table entry, possibly shifted. Completing the square writes a quadratic with complex roots as $(s - a)^{2} + b^{2}$, where $a \pm bi$ are its roots: the same $\alpha \pm \beta i$ that appeared in complex roots.
Worked example: Distinct linear factors
Find $\mathcal{L}^{-1}\left\{\frac{3s + 7}{(s + 1)(s + 3)}\right\}$.
- Write $\frac{A}{s + 1} + \frac{B}{s + 3}$.
- Cover-up at $s = -1$: $A = \frac{-3 + 7}{-1 + 3} = 2$. At $s = -3$: $B = \frac{-9 + 7}{-3 + 1} = 1$.
- Invert each piece.
Answer: $2e^{-t} + e^{-3t}$
Worked example: Completing the square
Find $\mathcal{L}^{-1}\left\{\frac{2s + 7}{s^{2} + 4s + 13}\right\}$.
- $s^{2} + 4s + 13 = (s + 2)^{2} + 9$.
- Rewrite the numerator around $s + 2$: $2s + 7 = 2(s + 2) + 3$.
- $F$ $= 2\cdot\frac{s + 2}{(s + 2)^{2} + 9}$ $+ \frac{3}{(s + 2)^{2} + 9}$, and the second piece is exactly $\mathcal{L}\{e^{-2t}\sin 3t\}$.
Answer: $e^{-2t}(2\cos 3t + \sin 3t)$
Worked example: A repeated factor
Find $\mathcal{L}^{-1}\left\{\frac{5}{(s - 2)^{3}}\right\}$.
- $\mathcal{L}\{t^{2}e^{2t}\} = \frac{2!}{(s - 2)^{3}}$.
- So $\frac{5}{(s - 2)^{3}} = \frac{5}{2}\cdot\frac{2}{(s - 2)^{3}}$.
Answer: $\frac{5}{2}t^{2}e^{2t}$
Common mistake
Completing the square but forgetting to rewrite the numerator: $\frac{2s + 7}{(s + 2)^{2} + 9}$ is not $2e^{-2t}\cos 3t + 7(\dots)$, because the cosine entry needs $s + 2$ upstairs, not $s$. Shift the numerator to $2(s + 2) + 3$ first, and divide the leftover constant by $b$ for the sine. Check: transforming your answer should give back $F$.
Why does the cover-up method work? If $F = \frac{A}{s + 1} + \frac{B}{s + 3}$, multiply both sides by $s + 1$ and set $s = -1$: the $B$ term vanishes and $A$ is what remains of $F$ with the factor $s + 1$ covered. For a repeated factor such as $(s - 2)^{3}$, you need every power: $\frac{A}{s - 2} + \frac{B}{(s - 2)^{2}} + \frac{C}{(s - 2)^{3}}$. A quick check of any answer: $f(0)$ should equal $\lim_{s\to\infty} sF(s)$; in the first example both are 3.