Question

integrate with respect to log x/x^2​

Answer 1

Given Question ;-

$\tt \:   \longrightarrow \:Evaluate \: \int \: \dfrac{logx}{ {x}^{2} } dx$

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$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

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Integration by Parts

Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways.

✏️See the rule:

✏️ ∫u v dx = u∫v dx −∫u' (∫v dx) dx

• u is the function u(x)

• v is the function v(x)

• u' is the derivative of the function u(x)

For integration by parts , the ILATE rule is used to choose u and v.

where,

• I - Inverse trigonometric functions

• L -Logarithmic functions

• A - Arithmetic and Algebraic functions

• T - Trigonometric functions

• E - Exponential functions

The alphabet which comes first is choosen as u and other as v.

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$\boxed{\tt \:   \longrightarrow \:\dfrac{d}{dx}logx = \dfrac{1}{x} }$

$\boxed{\tt \:   \longrightarrow \:\int \: {x}^{n} dx = \dfrac{ {x}^{ n+ 1} }{ n+1 } + c}$

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$\large\underline\purple{\bold{Solution :- }}$

$\tt \:   \longrightarrow \: \int \: \dfrac{1}{ {x}^{2} } log(x) dx$

$\tt \: = log(x)\int\dfrac{1}{ {x}^{2} } dx \: - \int \bigg(\dfrac{d}{dx}log(x)\int\dfrac{1}{ {x}^{2} dx} \bigg)dx$

$\tt \:  = log(x)\int {x}^{ - 2} dx - \int \bigg( \dfrac{1}{x} \int {x}^{ - 2} dx\bigg)dx$

$\tt \:  = log(x)\dfrac{ {x}^{ - 2 + 1} }{ - 2 + 1} - \int(\dfrac{1}{x} \times \dfrac{ {x}^{ - 2 + 1} }{ - 2 + 1} )dx$

$\tt \:  = - \: \dfrac{log(x)}{x} + \int\dfrac{1}{x} \times \dfrac{1}{x} dx$

$\tt \:  = - \: \dfrac{log(x)}{x} + \int \: {x}^{ - 2} dx$

$\tt \:  = - \: \dfrac{log(x)}{x} + \dfrac{ {x}^{ - 2 + 1} }{ - 2 + 1} + c$

$\tt \:  = - \: \dfrac{log(x)}{x} + \dfrac{ {x}^{ - 1} }{ - 1} + c$

$\tt \:  = - \: \dfrac{log(x)}{x} - \dfrac{ 1 }{ x} + c$

$\tt \:  = - \dfrac{1}{x} ( log(x) + 1) + c$

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