If In - -10ln xn-x dx- n - N- then

√(x) = t I_n = 2^n+1∫^1_0 (lnt)^n dt Let J_n = ∫^1_0 (lnt)^n dt J_n = [t(lnt)^n]^1_0 n ∫^1_0(lnt)^n 1 dt J_n = n J _n 1 ⇒I_n/2^n+1 = n. I_n 1/2^n⇒ I_n = ...

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Byju's Answer

Standard XIII

Mathematics

Integration by Parts

If In = ∫10ln...

Question

If $I_{n} = \int_{0}^{1} \frac{(l n x)^{n}}{\sqrt{x}} d x, n ϵ N,$ then

I_{n}

has finite value for all natural values of '

n

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\frac{I_{2}}{I_{1}} + \frac{I_{3}}{I_{2}} + \frac{I_{4}}{I_{3}} = - 4

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I_{4} = - 9

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I_{8} = 40331

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Solution

The correct option is A $I_{n}$ has finite value for all natural values of ' $n$ '
$\sqrt{x} = t$
$I_{n} = 2^{n + 1} \int_{0}^{1} (l n t)^{n} d t$
Let $J_{n} = \int_{0}^{1} (l n t)^{n} d t$
$J_{n} = [t (l n t)^{n}]_{0}^{1} - n \int_{0}^{1} (l n t)^{n - 1} d t$
$J_{n} = - n J_{n - 1}$
$\Rightarrow \frac{I_{n}}{2^{n + 1}} = - n . \frac{I_{n - 1}}{2^{n}} \Rightarrow I_{n} = - 2 n I_{n - 1}$
Now $I_{1} = \int_{0}^{1} \frac{l n x}{\sqrt{x}} d x = - 4$ , which is finite $\Rightarrow I_{n}$ is finite.

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If In=∫10(ln x)n√x dx,nϵN, then

The correct option is A In has finite value for all natural values of 'n'√x=t In=2n+1∫10(lnt)ndt Let Jn=∫10(lnt)n dt Jn=[t(lnt)n]10−n∫10(lnt)n−1dt Jn=−n Jn−1 ⇒In2n+1=−n.In−12n⇒In=−2n In−1 Now I1=∫10ln x√x dx=−4, which is finite ⇒In is finite.

If $I_{n} = \int_{0}^{1} \frac{(l n x)^{n}}{\sqrt{x}} d x, n ϵ N,$ then