Q50 / 104 · NTA Abhyas (JEE Main) 2020 · Mathematics · Indefinite Integration

If $I_{n} = \int {(\ln x)}^{n} d x$ , then $I_{10} + 10 I_{9}$ is equal to (where $C$ is the constant of integration)

Correct answer
A
$x {(\ln x)}^{10}$ $+ C$
B
$10 {(\ln x)}^{9}$ $+ C$
C
$9 {(\ln x)}^{10}$ $+ C$
D
$x {(\ln x)}^{9}$ $+ C$

Solution

I_{n} = \int {(\ln x)}^{n} d x

I_{n} = x {(\ln x)}^{n} - \int \frac{x n {(\ln x)}^{n - 1}}{x} d x

Ι_{n} = x {(\ln x)}^{n} - {nΙ}_{n - 1}

I_{n} + n I_{n - 1} = x {(\ln x)}^{n} + C

I_{10} + 10 I_{9} = x {(\ln x)}^{10}

+ C

Hence, the correct answer is option: 'A'.

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