Answer
$x^{2}+2x+3+\displaystyle \frac{1}{x+1}$
Work Step by Step
$\begin{array}{ccccccccccc}
& &x^{2} & +2x & +3 & \\
& &--&-- &--& \\
x+1&) & x^{3} & +3x^{2} & +5x& +4 & & \\
& & x^{3} & +x^{2 }& & & \color{red}{\leftarrow \small{x^{2}(x+1) } } \\
& &--&-- & & & \color{red}{ \small{subtract}} \\
& & & 2x^{2} & +5x & +4 & \\
& & & 2x^{2} & +2x& & \color{red}{\leftarrow \small{2x(x+1) } }\\
& & &--&-- & &\color{red}{ \small{subtract}} \\
& & & & 3x & +4 & \\
& & & & 3x & +3 & \color{red}{\leftarrow \small{3(x+1) } }\\
& & & &-- & -- &\color{red}{ \small{subtract}} \\
& & & & & 1 &
\end{array}$
Quotient = $x^{2}+2x+3$
Remainder = $1$.
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{ x^{3}+3x^{2}+5x+4}{x+1}$ = $x^{2}+2x+3+\displaystyle \frac{1}{x+1}$
