Edappally Thoppil, Kochi, India - 682021.
Details verified of Anuraj Raju✕
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Malayalam Mother Tongue (Native)
English Proficient
Kerala University 2012
BSC Mathematics
MG University 2015
Master of Computer Applications (M.C.A.)
Edappally Thoppil, Kochi, India - 682021
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Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
10
Board
State, ISC/ICSE, CBSE
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics, Computer Science
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
9
Board
IGCSE, International Baccalaureate, CBSE, State, ICSE
IB Subjects taught
Mathematics, Computers
CBSE Subjects taught
Computer Practices, Information and Comunication Technology
ICSE Subjects taught
Computer Application, Mathematics
IGCSE Subjects taught
Mathematics, Information Technology
Taught in School or College
No
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
9
Board
CBSE, State, ICSE
CBSE Subjects taught
Mathematics, Computer Practices, Information and Comunication Technology
ICSE Subjects taught
Computer Application, Mathematics
Taught in School or College
No
State Syllabus Subjects taught
Mathematics
1. Which school boards of Class 12 do you teach for?
State, ISC/ICSE and CBSE
2. Have you ever taught in any School or College?
Yes
3. Which classes do you teach?
I teach Class 10 Tuition, Class 12 Tuition and Class 9 Tuition Classes.
4. Do you provide a demo class?
Yes, I provide a free demo class.
5. How many years of experience do you have?
I have been teaching for 10 years.
Answered on 21/06/2020 Learn CBSE/Class 11
∫log x dx = x log x -x
Proof : Using integration by parts,
∫udv = uv - ∫vdu
In ∫ log x dx,
take, u=logx => du= (1/x) . dx
∫dv=∫dx => v=x
Now substituting,
∫log x = logx (x) - ∫x . 1/x . dx = logx (x) - ∫dx= x log x -x + C
where C is constant.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
10
Board
State, ISC/ICSE, CBSE
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics, Computer Science
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
9
Board
IGCSE, International Baccalaureate, CBSE, State, ICSE
IB Subjects taught
Mathematics, Computers
CBSE Subjects taught
Computer Practices, Information and Comunication Technology
ICSE Subjects taught
Computer Application, Mathematics
IGCSE Subjects taught
Mathematics, Information Technology
Taught in School or College
No
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
9
Board
CBSE, State, ICSE
CBSE Subjects taught
Mathematics, Computer Practices, Information and Comunication Technology
ICSE Subjects taught
Computer Application, Mathematics
Taught in School or College
No
State Syllabus Subjects taught
Mathematics
Answered on 21/06/2020 Learn CBSE/Class 11
∫log x dx = x log x -x
Proof : Using integration by parts,
∫udv = uv - ∫vdu
In ∫ log x dx,
take, u=logx => du= (1/x) . dx
∫dv=∫dx => v=x
Now substituting,
∫log x = logx (x) - ∫x . 1/x . dx = logx (x) - ∫dx= x log x -x + C
where C is constant.
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