# 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 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**

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 .

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) ^{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 = e^{z}

Now putting the value in equation (1)

we get, I = ∫ e^{z} ⋅z^{2}⋅e^{z} dz

or, I = ∫ e^{2z} ⋅ z^{2} dz

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

I = z^{2} ∫ e^{2z} 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= e^{2z} ⋅ \( \frac{z^2}{2}- \) ∫ e^{2z} z dz

once again apply the LIATE rule,

or, I = z^{2}. e^{2z}/2 – [ z ∫ e^{2z} dz- ∫ {dz/dz ∫ e^{2z} dz} dz

or, I= z^{2}. e^{2z}/2 – [ z⋅ e^{2z}/2 – e^{z}/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} ⋅ x^{2} – x^{2} ⋅ (log x) + \( \frac{1}{2} \) x^{2} + 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 + y ^{2} ).**

Ans. Let, I = ∫ log( 1 + y^{2} ) dy

Now using ILATE rule and we get,

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

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

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

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

Now separate the second part and get,

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

or, I = y ⋅ log( 1 + y^{2} ) – 2 [ x – tan^{-1}y ] + 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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