Question

∫log x dx,X belongs to (0,∞).

Answer 1

$\displaystyle \int log(x) dx$

We will integrate the above expression by parts i.e. u∫v dx −∫u' (∫v dx) dx

$\implies\displaystyle \int 1 \times log(x) dx$

Let , u = log x and v = 1

$\implies\displaystyle log(x) \int 1 \: dx - \int (\dfrac{d( log(x) )}{dx} \times \int 1dx)dx$

$\implies\displaystyle log(x) x\: - \int (\dfrac{1}{x} \times x)dx$

$\implies\displaystyle \: x log(x) \: - \int ({1})dx$

$\implies\displaystyle \: x log(x) \: - x + c$

where c is constant.

Answer :

$\displaystyle \: x log(x) \: - x + c$

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Learn more :-

∫ 1 dx = x + C

∫ sin x dx = – cos x + C

∫ cos x dx = sin x + C

∫ sec2 dx = tan x + C

∫ csc2 dx = -cot x + C

∫ sec x (tan x) dx = sec x + C

∫ csc x ( cot x) dx = – csc x + C

∫ (1/x) dx = ln |x| + C

∫ ex dx = ex+ C

∫ ax dx = (ax/ln a) + C

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Answer 2

Given ,

The function is log(x)

Integrating function wrt to x , we get

$\tt \mapsto \int{ log(x) \: dx}$

$\tt \mapsto \int{ log(x) \times 1 \: dx}$

We know that ,

$\tt \int uv \: \: dx = u \int{ v } \: \: dx - \int{ \bigg\{ \frac{du}{dx} \int{v \: \: dx} \bigg\}} \: \: dx$

Where ,

u = first function

v = second function

Thus ,

Let , log(x) as a first function and 1 as a second function

$\tt \mapsto log(x) \int{ 1 \: dx} - \int{ \bigg\{ \frac{ d \{\log(x) \}}{dx} \int{1 \: \: dx} \bigg\}} \: \: dx$

$\tt \mapsto xlog(x) \int{ \frac{1}{x} \times x} \: \: dx$

$\tt \mapsto xlog(x) - \int{ 1} \: \: dx$

$\tt \mapsto xlog(x) - x + c$

Remmember :

$\tt \int{ {x}^{n} } \: \: dx = \frac{ {(x)}^{n + 1} }{n + 1}$

$\tt \frac{d \{log(x) \}}{dx} = \frac{1}{x}$

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