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$I_{n}=\int \left ( log\, x \right )^{n}dx,$ then $I_{n}+nI_{n-1}$ =
Option: 1
$n\left ( log\: X \right )^{n}$

Option: 2
$\left (n\: log\: X \right )^{n-1}$

Option: 3
$\left (log\: X \right )^{n-1}$

Option: 4
$\left (n\, log\: X \right )^{n}$

Answers (1)

Integration By PARTS -

Let $u$ and $v$ are two functions then

$\int u\cdot vdx=u\int vdx-\int \left ( \frac{du}{dx}\int vdx \right )dx$

- wherein

Where $u$ is the Ist function $v$ is he IInd function

Rule for integration by parts -

Take Ist function $(u)$ as according I L A T E

- wherein

Where ,

I : Inverse

L : Logarithmic

A : Algebraic

T : Trignometric

E : Exponential

$I_{n}=\int (log\: x)^{n}dx$

$\therefore I_{n-1}=\int (log\: x)^{n-1}dx$

Now $I_{n}=\int (log\: x)^{n}1.dx$

$= (log\: x)^{n}.x-n\int (log\: x)^{n-1}.\frac{1}{x}.x\; dx$

$I_{n}=n(log\: x)^{n}-n\: I_{n-1}$

$I_{n}+n\: I_{n-1}=n(log\: x)^{n}$

Posted by

Ramraj Saini

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