Question

Let $I_n(x)=\underset{0}{\overset{x}{\int }}\frac{1}{{\left(t^2+5\right)}^n}dt,n=1,2,3,\dots$ Then

• A. $50I_6-9I_5=x{I}_5^{\prime }$
• B. $50I_6-11I_5=x{I}_5^{\prime }$
• C. $50I_6-9I_5=I_5^{\prime }$
• D. $50I_6-11I_5=I_5^{\prime }$

Answer 1

Answer: (A)

$I_n(x)=\underset{0}{\overset{x}{\int }}\frac{dt}{{\left(t^2+5\right)}^n}$
Applying integral by parts
$I_n(x)={\left[\frac{t}{{\left(t^2+5\right)}^n}\right]}_0^x-\underset{0}{\overset{x}{\int }}n{\left(t^2+5\right)}^{-n-1}\cdot 2t^2$
$I_n(x)=\frac{x}{{\left(x^2+5\right)}^n}+2n\underset{0}{\overset{x}{\int }}\frac{t^2}{{\left(t^2+5\right)}^{n+1}}dt$
$I_n(x)=\frac{x}{{\left(x^2+5\right)}^n}+2n\underset{0}{\overset{x}{\int }}\frac{\left(t^2+5\right)-5}{{\left(t^2+5\right)}^{n+1}}dt$
$I_n(x)=\frac{x}{{\left(x^2+5\right)}^n}+2n{I}_n(x)-10n{I}_{n+1}(x)$
$10n{I}_{n+1}(x)+(1-2n)I_n(x)=\frac{x}{{\left(x^2+5\right)}^n}$
$Putn=5$