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 $n=5$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 7.2 definite integral→STD 12 - 7.2 definite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Choose the correct answer from the given four options.
Let A and B be two events such that $\text{P}(\text{A})=\frac{3}{8},\text{P}({\text{B}})=\frac{5}{8}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{4}.$Then $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}\Big(\frac{\text{A'}}{\text{B}}\Big)$ is equal to:

$\int_{ - 3}^3 {\frac{{{x^2}\sin 2x}}{{{x^2} + 1}}\,dx = } $

If $\alpha,\beta$ and $\gamma$ are the angles which a half ray makes with the positive direction of the axes, then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ is equal to:

Choose the correct answers from the given four options:
The derivative of $\cos^{-1}(2\text{x}^2-1)$ w.r.t. $\cos^{-1}\text{x}$ is:

$\left[ {\int\limits_0^2 {\sqrt {x + \sqrt {x + \sqrt {x + .....\infty } } \,dx} } } \right]$ is equal to (where $[·]$ is $G.I.F.$)

Choose the correct answer from the given four options.
For any vector $\vec{\text{a}},$ the value of $(\vec{\text{a}}\times\hat{\text{i}})^2+(\vec{\text{a}}\times\hat{\text{j}})^2+(\vec{\text{a}}\times\hat{\text{k}})^2$ is:

Let $y=y(x)$ be the solution curve of the differential equation

$\sin \left(2 x^{2}\right) \log _{c}\left(\tan x^{2}\right) d y+\left(4 x y-4 \sqrt{2} x \sin \left(x^{2}-\frac{\pi}{4}\right)\right) d x=0$

$0 < x < \sqrt{\frac{\pi}{2}}$, which passes through the point $\left(\sqrt{\frac{\pi}{6}}, 1\right)$. Then $\left|y\left(\sqrt{\frac{\pi}{3}}\right)\right|$ is equal to $.....$

The area of the bounded region enclosed by the curve $y=3-\left|x-\frac{1}{2}\right|-|x+1|$ and the $x-$axis is

If $A = \left[ {\begin{array}{*{20}{c}}i&0\\0&{i/2}\end{array}} \right]$ $(i = \sqrt { - 1} ),$then ${A^{ - 1}}$=

A pair of fair dice is rolled together till a sum of either $5$ or $7$ is obtained. Then the probability that $5$ comes before $7$ is