Answer
$ \frac{a^2+1}{b(a+2)}$.
Work Step by Step
The given expression is
$\Rightarrow \frac{a^2b+b}{3a^2-4a-20}\cdot \frac{a^2+5a}{2a^2+11a+5}\div \frac{ab^2}{6a^2-17a-10}$
Invert the divisor and multiply.
$\Rightarrow \frac{a^2b+b}{3a^2-4a-20}\cdot \frac{a^2+5a}{2a^2+11a+5}\cdot \frac{6a^2-17a-10}{ab^2}$
Factor each numerator and denominator as shown below.
$\Rightarrow a^2b+b$
Factor out $b$.
$\Rightarrow b(a^2+1)$
$\Rightarrow 3a^2-4a-20$
Rewrite the middle term $-4a$ as $-10a+6a$.
$\Rightarrow 3a^2-10a+6a-20$
Group terms.
$\Rightarrow (3a^2-10a)+(6a-20)$
Factor each group.
$\Rightarrow a(3a-10)+2(3a-10)$
Factor out $(3a-10)$.
$\Rightarrow (3a-10)(a+2)$
$\Rightarrow a^2+5a$
Factor out $a$.
$\Rightarrow a(a+5)$
$\Rightarrow 2a^2+11a+5$
Rewrite the middle term $11a$ as $10a+1a$.
$\Rightarrow 2a^2+10a+1a+5$
Group terms.
$\Rightarrow (2a^2+10a)+(1a+5)$
Factor each group.
$\Rightarrow 2a(a+5)+1(a+5)$
Factor out $(a+5)$.
$\Rightarrow (a+5)(2a+1)$
$\Rightarrow 6a^2-17a-10$
Rewrite the middle term $-17a$ as $-20a+3a$.
$\Rightarrow 6a^2-20a+3a-10$
Group terms.
$\Rightarrow (6a^2-20a)+(3a-10)$
Factor each group.
$\Rightarrow 2a(3a-10)+1(3a-10)$
Factor out $(3a-10)$.
$\Rightarrow (3a-10)(2a+1)$
Substitute the factors into the given expression.
$\Rightarrow \frac{b(a^2+1)}{(3a-10)(a+2)}\cdot \frac{a(a+5)}{(a+5)(2a+1)}\cdot \frac{(3a-10)(2a+1)}{ab^2}$
Cancel common terms.
$\Rightarrow \frac{a^2+1}{b(a+2)}$.
