Question

integration of logx​

Answer 1

Here we use Integration by parts formula :

$\boxed{ \bold{\int \: uv \: dx = u \int \: v \: dx - \int \: ( \frac{du}{dx} \int \: v \: dx)dx}}$

•This formula states that :

= First function u multiplied by integral of 2nd function v - integral of [ d.c of 1st function multiplied by integral of 2nd already written before ]

Here we choose

1st and 2nd function by

◆ILATE

= I( Inverse trigonometry)

=L (logarithmic function)

=A ( algebraic function)

= T( Trigonometry function

=E(Exponential function )

# If a ques have two function in that of both

log and trigonometry

then we take

1st function as Log

and

2nd function as trigonometry .

Soln refers to the attachment

Answer 2

Question :-

$\implies \sf{ \int \: log(x) dx \:}$

Answer :-

$\implies \: \boxed{\sf{ \pink{ x \: log(x)} \red{ - x + c}} \: }$

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Solution :-

Here we used integration by parts .

$\rightarrow \sf{ \int \: 1 \: \times log(x) dx}$

By using the formula of integration by parts ,

$\footnotesize \pink{ \sf{log(x) \int 1 dx} - \int \green{ \bigg( \frac{d}{dx} log(x) \int 1 \: dx \bigg) } dx} \\ \\ \because \green{ \sf{ \frac{d}{dx} log(x)} = {\frac{1}{x} }} \\ \\ \rightarrow \red{\sf{ log(x) \times x} - \pink{ \int \frac{1}{x} \times x \: dx} }\\ \\ \rightarrow \green{ \sf{x \: log(x) } - \red{ \int \: 1 \: dx} }\\ \\ \rightarrow \sf{ \pink{ x \: log(x) - x + c}}$

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