Explanation

(b)
Put $x + 1 = t^{2}$
$\Rightarrow \ dx = 2t\ dt$
$\Rightarrow \ x^{2} + 1 = \left( t^{2} - 1 \right)^{2} + 1$
$= t^{4} - 2t^{2} + 2$
$\therefore$ Given integral
$= \int_{}^{}{\left( t^{4} - 2t^{2} + 2 \right)t.2t\ dt}$
$= 2\int_{}^{}{\left( t^{6} - 2t^{4} + 2t^{2} \right)dt}$
$= 2\left\lbrack \frac{t^{7}}{7} - 2\frac{t^{5}}{5} + 2\frac{t^{3}}{3} \right\rbrack + c$
$= 2\left\lbrack \frac{(x + 1)^{\frac{7}{2}}}{7} - 2\frac{(x + 1)^{\frac{5}{2}}}{5} + 2\frac{(x + 1)^{\frac{3}{2}}}{3} \right\rbrack + c$