If In= displaystyle ∫ (ln x)ndx , then I10+10I9 is equal to (where C is the constant of integration)

Question

If In= displaystyle ∫ (ln x)ndx , then I10+10I9 is equal to (where C is the constant of integration) (A) x(ln x)10 +C (B) 10(ln x)9 +C (C) 9(ln x)10 +C (D) x(ln x)9 +C

Tardigrade Admin · Accepted Answer

In= displaystyle ∫ (ln x)ndx In=x(ln x)n- displaystyle ∫ (x n (ln â¡ x)n - 1/x)dx În=x(ln x)n-(nÎ)n - 1 In+nIn - 1=x(ln x)n+C I10+10I9=x(ln x)10 +C

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

$x (l n x)^{10}$ $+ C$

$10 (l n x)^{9}$ $+ C$

$9 (l n x)^{10}$ $+ C$

$x (l n x)^{9}$ $+ C$

$I_{n} = \int (l n x)^{n} d x$
$I_{n} = x (l n x)^{n} - \int \frac{x n ( l n x ) ^{n - 1}}{x} d x$
$I_{n} = x (l n x)^{n} - (n I)_{n - 1}$
$I_{n} + n I_{n - 1} = x (l n x)^{n} + C$
$I_{10} + 10 I_{9} = x (l n x)^{10}$ $+ C$

Q. If In​=∫(lnx)ndx , then I10​+10I9​ is equal to (where C is the constant of integration)

x(lnx)10 +C

10(lnx)9 +C

9(lnx)10 +C

x(lnx)9 +C