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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$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad k \quad, \quad x=0$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1} ,\,\,\, x>0$
એ $x=0$ આગળ સતત હોય તો $\frac{1}{a}+\frac{1}{b}+\frac{4}{k}$ ની કિમંત મેળવો.
(કે જ્યાં $\mathrm{C}$ એ સંકલન અચળાંક છે )