Applying integral by parts
$I_{n}(x)=\left[\frac{t}{\left(t^{2}+5\right)^{ n }}\right]_{0}^{ x }-\int_{0}^{ x } n \left( t ^{2}+5\right)^{- n -1} \cdot 2 t ^{2}$
$I _{ n }( x )=\frac{ x }{\left( x ^{2}+5\right)^{ n }}+2 n \int_{0}^{ x } \frac{ t ^{2}}{\left( t ^{2}+5\right)^{ n +1}} dt$
$I _{ n }( x )=\frac{ x }{\left( x ^{2}+5\right)^{ n }}+2 n \int_{0}^{ x } \frac{\left( t ^{2}+5\right)-5}{\left( t ^{2}+5\right)^{ n +1}} dt$
$I _{ n }( x )=\frac{ x }{\left( x ^{2}+5\right)^{ n }}+2 n I _{ n }( x )-10 n I _{ n +1}( x )$
$10 n I _{ n +1}( x )+(1-2 n ) I _{ n }( x )=\frac{ x }{\left( x ^{2}+5\right)^{ n }}$
Put $n=5$
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$(\alpha + p)^{m - 1} + (\alpha + p)^{m - 2} (\alpha + q) + (\alpha + p)^{m - 3} (\alpha + q)^2 + ...... (\alpha + q)^{m - 1}$
where $\alpha \ne - q$ and $p \ne q$ is :
$7 \times 8,10 \times 10,13 \times 12,16 \times 14, \ldots .$ is ....... .
$\quad \quad \quad \quad \quad 5 x+1,\quad \quad \quad \quad \quad x \leq 2$, then