Let I_ n (x)= _ 0 ^ x 1 (t^ 2 +5 )^ n d t, n=1,2,3, . Then - Vidyadip

MCQ

Let $I_{n}(x)=\int_{0}^{x} \frac{1}{\left(t^{2}+5\right)^{n}} d t, n=1,2,3, \ldots .$ Then

✓
$50\,I _{6}-9\,I _{5}= xI _{5}^{\prime}$
B
$50\,I _{6}-11\,I _{5}= xI _{5}^{\prime}$
C
$50\,I _{6}-9\,I _{5}= I _{5}^{\prime}$
D
$50\,I _{6}-11\,I _{5}= I _{5}^{\prime}$

✓

Answer

Correct option: A.

$50\,I _{6}-9\,I _{5}= xI _{5}^{\prime}$

a
$I_{n}(x)=\int_{0}^{ x } \frac{ dt }{\left( t ^{2}+5\right)^{ n }}$

Applying integral by parts

$I_{n}(x)=\left[\frac{t}{\left(t^{2}+5\right)^{ n }}\right]_{0}^{ x }-\int_{0}^{ x } n \left( t ^{2}+5\right)^{- n -1} \cdot 2 t ^{2}$

$I _{ n }( x )=\frac{ x }{\left( x ^{2}+5\right)^{ n }}+2 n \int_{0}^{ x } \frac{ t ^{2}}{\left( t ^{2}+5\right)^{ n +1}} dt$

$I _{ n }( x )=\frac{ x }{\left( x ^{2}+5\right)^{ n }}+2 n \int_{0}^{ x } \frac{\left( t ^{2}+5\right)-5}{\left( t ^{2}+5\right)^{ n +1}} dt$

$I _{ n }( x )=\frac{ x }{\left( x ^{2}+5\right)^{ n }}+2 n I _{ n }( x )-10 n I _{ n +1}( x )$

$10 n I _{ n +1}( x )+(1-2 n ) I _{ n }( x )=\frac{ x }{\left( x ^{2}+5\right)^{ n }}$

Put $n=5$

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