INTEGRATION OF LOG X | integration of logx

Integration of log x in pdf | Integral of log x

Hello students I am Bijoy and welcome to my educational forum. Today we will deal with the Integration of logx. There we see various types of integration on logx, & also will see something about on the LIATE or ILATE rule of by parts method.

integration pf log x
                                    Integration of log x

Integration of logx

Lets start, the evaluating the integration of log x

Let, I = ​\( \int log|x| dx \)

In previous section of integration we have learnt basic integration, substitution method and by parts method for integrating the functions. but here students if try to observe then you can understand easily that we can not integrate logx by basic integration or substitution, so we have only one option that to integrate log x, with by parts that to.

Now I want just recap to all of you the by parts method or rule, like LIATE or ILATE rule, just remind that.

so, lets start                                          I = ​\( \int log|x| dx \)

students if you observe , you can see two function logx and 1, now we can apply LIATE rule and we get,

or, I = ​\( logx \int dx – \int (d(logx)/dx \int dx)dx \)

or, I = ​\( logx .(x) – \int (\frac {1}{x}).x. dx \)

or, I = ​\( x.logx – \int dx \)

or, I = x . logx – x + c      [ where ‘c’ is integral constant }

Hence, \( \int log|x| dx \)​= x.logx – x + c  = x ( logx – 1 ) + c

Now we will try to evaluate another type of integration that is integration of log ax.

so, our integration

I = ​\( \int log|ax|dx \)

students same as the above, two functions logax and 1, now applying LIATE and get,

or,   I =  ​\( log|ax| \int dx – \int(\frac{dlog|ax|}{dx} \int dx)\int dx \)

or, I =  ​\( log|ax|.x – \int (\frac{a.x}{x})dx \)

or, I = x log|ax| – x + c                          [ where ‘c’ is integral constant ]

So required answer equal to x log|ax| + c

# Evaluate the integration of 1/xlogx
integration of log x
                                                                                 integration of1/ log x

Our required integration, I = ​\( \int(\frac{1}{xlogx}) dx \)​    ———- (1)

This is the one of the simple on logx problems, in this case you have to a simple substitution on log x.

so, assume that log x = z, now differentiating on both side with the respect to ‘x’ and we get,

\( \frac{d(logx)}{dx} = \frac{dz}{dx} \)

or, ​\( \frac{1}{x} dx = dz \)

Now putting the value in equation no (1) and we get,

or, I = ​\( \int \frac{1}{x} dx *logx \)

or, I = ​\( \int \frac{dz}{z} \)

or, I = log|z| + c                                              [ where ‘c’ is integral constant ]

or. I = log|log|x|| + c

Hence, integration of 1/xlogx equal to log|log x|| + c

Students, now try something with good question. Find out the value of integration of (log x)2 dx .
integration of logx
                                                                     integration of (log x)2

so we have to find the value of, I = ​\( \int (logx)^2 dx \)

Students, if try to observe then you can see the two functions one is (logx)2 and another one is 1.

So, now applying LIATE rule and we get,

I = (logx)2 ∫1.dx – ​\( \int [\frac{d(logx)^2}{dx} \int 1.dx]dx \)

or, I = x ⋅ (log x)2 – ​\( \int \frac{2.logx}{x}.x dx \)

or, I = x⋅ (log x)2 – 2 ∫ (logx) dx

Now using by parts once again in second part and we get,

or, I = x⋅ (log x)2 -2x logx + 2x + c                [ Where ‘c’ is integral constant]

∴ I = x (logx)2  – 2 x (logx – 1) + c

Ex. Evaluate the integration of log(1/x).

Ans.                   Let, I = ​\( \int log(\frac{1}{x}) dx \)​     ———— (1)

or, I = log (1/x) ∫dx – ​\( \int [(\frac{dlog(1/x)}{dx})\int dx]dx \)

or, I = x⋅log(1/x) – ​\( \int( -\frac{1}{x^2} x.x dx \)

or, I = x⋅log(1/x) + ∫ dx

∴ I = x⋅log(1/x) + x + c               [ where ‘c’ is integral constant ]

Hence, the integration of log(1/x) equal to x⋅log(1/x) + x + c.

Ex.2

What is the value of ​\( \int [\frac{1}{log x}- \frac{1}{(logx)^2}] \)​  ?

Ans. For solving that integration first we have to start from,

Let, I = ​\( \int \frac{1}{logx} dx \)​   ——— (1)

Now using by parts and we get,

or, I = ​\( \frac{1}{logx} \int dx – \int [(\frac{d(\frac{1}{logx})}{dx})\int dx]dx \)

or, I = ​\( x \frac{1}{log x} – \int[-\frac{1}{(log x)^2}* \frac{x}{x} dx] \)

or, I = ​\( \frac{x}{logx} +\int \frac{dx}{(log x)^2} + c \)

[ where c is integral constant ]

Now applying the side changing rule on second part of given integration and we get,

or, I – \(\int \frac{dx}{(log x)^2} + c = x⋅\frac{1}{log x}\)

Now put the value of I, from equation no (1) we get,

\( \int \frac{dx}{log x} – \int \frac{dx}{(logx)^2}= x\frac{1}{log x} + c \)

Ex.3

What is the value of integration of x⋅(log x)?

integration of log x in pdf
                                                                      integration of x.(log x)2

Ans.                                 Let,   I = ∫ x ⋅ (log x)2 dx   —————- (1)

Students first we use substitution method .

now assume,  log x = z and differentiating on both side with respect to x.

and we get,  1/x dx = dz, and dx = x⋅dz

again  log x = z , ∴ x = ez

Now putting the value in equation (1)

we get,   I = ∫ ez ⋅z2⋅ez dz

or, I = ∫ e2z ⋅ z2 dz

Now, by using by parts as LIATE rule, we get

I = z2 ∫ e2z dz – ​\( \int [\frac{dz^2}{dz} \int e^(2z) dz]dz \)

or, I = ​\( \frac{e^(2z) . z^2}{2} – \int \frac{2z}{2} * e^(2z) dz \)

or, I= e2z ⋅ ​\( \frac{z^2}{2}- \)​ ∫ e2z z dz

once again apply the LIATE rule,

or, I = z2. e2z/2 – [ z ∫ e2z dz- ∫ {dz/dz ∫  e2z dz} dz

or, I= z2. e2z/2 – [ z⋅ e2z/2 – ez/4] + c            [ where  c is integral constant ]

or, I = ​\( \frac{z^2}{2} (e^z)^2 – \frac{z}{2}(e^z)^2 + \frac{1}{4} (e^z)^2 + c \)

or, I = ​\( \frac{1}{2} \)​ [ (log x)2 ⋅ x2 – x2 ⋅ (log x) + \( \frac{1}{2} \) x2 + c ]

∴ I =  ​\( \frac{1}{2} x^2 \)​ [ (logx)2 – (logx) + 1/2 + c]

Hence, the integration of x ⋅(log x)2 equal to ​\( \frac{1}{2} x^2 \)​ [ (logx)2 – (logx) + 1/2 + c]

Ex. Find the value of integration of log (1 + y2 ).

Ans.                            Let,   I = ∫ log( 1  + y2 ) dy

Now using ILATE  rule and we get,

or, I = log( 1  + y2 ) ∫ dy  – ​\( \int [\frac{dlog(1+y^2)}{dy} \int dy] dy \)

or, I = y ⋅ log( 1  + y2 )  – ​\( \int \frac{2y}{1+y^2} ydy \)

or, I = y ⋅ log( 1  + y2 ) – ​\( 2\int \frac{y^2}{1+ y^2} dy \)

or, I = y ⋅ log( 1  + y2 ) – ​\( 2\int \frac{1+y^2-1}{1+y^2}dy \)

Now separate the second part and get,

or, I = y ⋅ log( 1  + y2 ) – ​\( 2[\int \frac{1+y^2}{1+y^2}dy – \int \frac{dy}{1+y^2}] \)

or, I = y ⋅ log( 1  + y2 ) – 2 [ x – tan-1y ] + c             [ where c is integral constant ]

This is your required answer, so students if you want to more illustration then please comment and thank you. Just visit on  INTEGRATION OF SECX | INTEGRATION OF SEC^2X.

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