JEE Main 2022 — Mathematics Calculus

Given,

$${I}_{n}(x)={\int }_{0}^{x}\frac{dt}{{({t}^{2}+5)}^{n}}$$In​(x)=∫0x​(t2+5)ndt​

Applying integration by parts we get,

$$\Rightarrow {I}_{n}(x)={[\frac{t}{{({t}^{2}+5)}^{n}}]}_{0}^{x}-{\int }_{0}^{x}n{({t}^{2}+5)}^{-n-1}\cdot 2{t}^{2}$$⇒In​(x)=[(t2+5)nt​]0x​−∫0x​n(t2+5)−n−1⋅2t2

$$\Rightarrow {I}_{n}(x)=\frac{x}{{({x}^{2}+5)}^{n}}+2n{\int }_{0}^{x}\frac{{t}^{2}}{{({t}^{2}+5)}^{n+1}}dt$$⇒In​(x)=(x2+5)nx​+2n∫0x​(t2+5)n+1t2​dt

$$\Rightarrow {I}_{n}(x)=\frac{x}{{({x}^{2}+5)}^{n}}+2n{\int }_{0}^{x}\frac{({t}^{2}+5)-5}{{({t}^{2}+5)}^{n+1}}dt$$⇒In​(x)=(x2+5)nx​+2n∫0x​(t2+5)n+1(t2+5)−5​dt

$$\Rightarrow {I}_{n}(x)=\frac{x}{{({x}^{2}+5)}^{n}}+2n{\int }_{0}^{x}\frac{dt}{{({t}^{2}+5)}^{n}}-10n{\int }_{0}^{x}\frac{dt}{{({t}^{2}+5)}^{n+1}}$$⇒In​(x)=(x2+5)nx​+2n∫0x​(t2+5)ndt​−10n∫0x​(t2+5)n+1dt​

$$\Rightarrow {I}_{n}(x)=\frac{x}{{({x}^{2}+5)}^{n}}+2n{I}_{n}(x)-10n{I}_{n+1}(x)$$⇒In​(x)=(x2+5)nx​+2nIn​(x)−10nIn+1​(x)

$$\Rightarrow 10n{I}_{n+1}(x)+(1-2n){I}_{n}(x)=\frac{x}{{({x}^{2}+5)}^{n}}$$⇒10nIn+1​(x)+(1−2n)In​(x)=(x2+5)nx​

$\Rightarrow 10n{I}_{n+1}(x)+(1-2n){I}_{n}(x)=x{{I}^{'}}_{n}$⇒10nIn+1​(x)+(1−2n)In​(x)=xI′n​ {where ${{I}^{'}}_{n}=\frac{1}{{({x}^{2}+5)}^{n}}$I′n​=(x2+5)n1​ we get by differentiating $\displaystyle {I}_{n}(x)={\int }_{0}^{x}\frac{dt}{{({t}^{2}+5)}^{n}}$In​(x)=∫0x​(t2+5)ndt​ }

Now put $n=5$n=5

We get, $50{I}_{6}-9{I}_{5}=x{I}_{5}^{'}$50I6​−9I5​=xI5′​