Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

Find the antiderivative of f(x)=ln(lnx)+(lnx)2 whose graph passes through (e,e)

Answer
VerifiedVerified
492k+ views
like imagedislike image
Hint: Antiderivative means integration and used for the opposite of the word derivative. Apply integration by parts wherever required which is given as
f(x)g(x)dx=f(x)g(x)dxf(x)g(x)dxdx
And use the following identities
 ddx(lnx)=1x,xndx=xn+1n+1

Complete step-by-step answer:
The word antiderivative is the substitution of the word ‘integration’ or opposite of derivative.
So, we have function f(x)=ln(lnx)+(lnx)2 where we need to determine antiderivative of f (x) if it’s passing through (e, e).
Now, let I be the antiderivative of f(x); hence, we can write
I=(ln(lnx)+(lnx)2)dx ……………… (i)
Now, we can substitute ‘lnx’ as ‘t’ for simplifying the given integral so, we have
lnx = t or
logex=t
As we know that logax=N can be written as x=aN. So, we get from the above relation as:
x=et
Now, differentiate it w.r.t ‘x’
ddx(x)=ddx(et)1=etdtdxdx=etdt...............(ii)
ddxex=ex
So, we can re-write equation (i) with respect to variable ‘t’ as
I=(lnt+t2)etdt
I=etlntdt+ett2dt I I2 …………………… (iii)
Let us calculate value of I2 with the help of integration by parts which is given as
f(x)g(x)dx=f(x)g(x)dxf(x)g(x)dxdx
So, I2 can be expressed in form of integration by parts by supposing f(x)=etand g(x)=(1t2) in the above equation by the ILATE rule, it can be given as:
ILATE stands for
I – Inverse trigonometry
L – Logarithm
A – Algebraic
T – Trigonometry
E – Exponential
It means ILATE consists of 5 different functions as mentioned above. So, the purpose of this rule is to decide the priority of the first function and second i.e. which function is going for the integration and which is going for the differentiation. So, according to the above order, we decide the priority of function (1 and 2) involved in the integration by parts. So, we can use above mentioned identity with the given integral I2
Hence, we get
I2=ett2dt=ett2dtddtett2dtdt
As we know
xndx=xn+1n+1,ddx(ex)=ex
Hence, on simplifying equation of I2 , we get
I2=et(1t)et(1t)dt
Or
I2=ettt+ettdt
Now, using the integration by parts again with the ettdt, we get
I2=ett+et1tdtddtet1tdtdt
Now, we know
1xdx=logex
ddx=logex
Hence, we get
I2=ett+etlogetetlogetdt+c.............(iv)
Now, we can put value of I2 from equation (iv) to equation (iii) and hence, we get
I=etlogetdt+ett+etloget+cetlogetdt
So, we can cancel the term etlogetdt from the above expression.
Hence, we get
I=ettt+etloget+c
Now, substitute the value of t as logex. Hence, we get
I=elogexlogex+elogexlogelogex+c
As, we know (a)logab=b. Hence, we get
I=xlogex+xloge(logex)+c
Hence, antiderivative of function f(x) is given as
I=xlogex+xloge(logex)+c
Or
y=xlogex+xloge(logex)+c………………. (v)
Now, we can find the value of ’c’ by using the information that the graph of antiderivative is passing through (e, e).
Hence, (e, e) will the satisfy the equation (v), so, we get
e=elogee+eloge(logee)+c
So, we know logee=1 and loge1=0. Hence we get
e = - e + e (o )+ c
c = 2e
Hence, equation of antiderivative of f(x) can be given from equation (v) ad
xlogex+xloge(logex)+2e

Note: One can integrate the term etlntdt as well for getting the answer. But we can not find the exact values of I1 and I2. So, integrate only one function to cancel out the other. It is the key point for getting the integration I and hence the solution as well.
One may get confused with the term antiderivative. He or she may calculate differentiation of the given function and then try to calculate the inverse of the derivative as well, which is wrong. Antiderivative used for the word integration. So, be clear with the terms as well for these kinds of questions.

Latest Vedantu courses for you
Grade 10 | MAHARASHTRABOARD | SCHOOL | English
Vedantu 10 Maharashtra Pro Lite (2025-26)
calendar iconAcademic year 2025-26
language iconENGLISH
book iconUnlimited access till final school exam
tick
School Full course for MAHARASHTRABOARD students
PhysicsPhysics
BiologyBiology
ChemistryChemistry
MathsMaths
₹36,600 (9% Off)
₹33,300 per year
Select and buy