Question

find a reduction formula integral of x^m /(log x)^n dx​

Answer 1

Answer:

answer is in the photo.........

Answer 2

Explanation:

$\tt \: Let, \: I_{m,n} = \int \dfrac{x {}^{m} }{(log_x) {}^{n} }.dx \\ \\ \longrightarrow \tt I_{m,n} = \int x {}^{m} (log_x) {}^{ - n} .dx \\ \\ \longrightarrow \tt I_{m,n} = \int (log_x) {}^{ - n} \int x {}^{m} dx - \int( - n)(log_x) {}^{ - n - 1} dx \\ \\ \longrightarrow \tt I_{m,n} = \int (log_x) {}^{ - n } \: \dfrac{x {}^{m + 1} }{m + 1} + \frac{n}{m + 1} \: I_m,(n + 1) \\ \\ \longrightarrow \boxed{\tt I_{m,n} = \int I_m,(n + 1) \:( \dfrac{n}{m + 1} ) + \dfrac{x {}^{m + 1} }{(m + 1)(log_x) {}^{n} } } \\ \\ \longrightarrow \tt I_m,(n + 1) = I_{mn} \: ( \dfrac{m + 1}{n} ) - \dfrac{x {}^{m + 1} }{(m + 1)} \times( \dfrac{m + 1}{n} ) \times \frac{1}{(log_x) {}^{n} } \\ \\ \longrightarrow \tt \:I_m,(n + 1) = I_{m,n }( \dfrac{m + 1}{n} ) - \dfrac{x {}^{m + 1} }{(n)(log_x) {}^{n} } \\ \\ \tt \: also,can \: write \: replace \: \red{n \rightarrow (n - 1).} \\ \\ \longrightarrow { \boxed{ \tt{I_m,(n + 1) = I_{m,n }( \dfrac{m + 1}{n - 1} ) - \dfrac{x {}^{m + 1} }{(n - 1)(log_x) {}^{n - 1} } }}}$

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Know to more:

• Power rule.

$\bull\tt \int x {}^{n} .dx \: \rightarrow \: \dfrac{x {}^{n + 1} }{n + 1} + c$

• Multiplication by constant (c).

$\bull\tt \int c \: f(x).dx \: \rightarrow \: c \int f(x).dx$

• Sum rule.

$\bull\tt \int (f + g)dx \rightarrow \: \int fdx \: + \: \int gdx$

• Difference rule.

$\bull \tt \int (f - g)dx \rightarrow \: \int fdx \: - \: \int gdx$