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What is the integration process of e^x/(1+x)

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such question is solved by using by parts but in this question by parts will not work in fact this not integrable
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Maths Solution

Integration by parts....assuming (e^x ) 2nd function and 1/(1+x) ist function
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M.Tech from IIT Bombay(Expert maths Faculty)

Apply ILATE rules (standing for Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential), Integration by parts.
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Master of Physics & have excellent skills of Mathematics problem solving

Use LIATE (Logarithmic-Inverse-Algebric- Trignometruc-Expoential Sequence) with u/v rule
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e^x/x+1
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to solve such type problem, you have to use this method ,if both the functions are directly integrable ,then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable . while doing this if you find this question in the form...
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to solve such type problem, you have to use this method ,if both the functions are directly integrable ,then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable . while doing this if you find this question in the form of the such function sinx/x, cosx/x ,1/logx , rootsinx , sin(x square ) exp.(x square) exp.(-x square) ,xtanx , and few more ..however these functions are differentiable but integration can not be found read less
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Mukesh Rajbhar

Byparts rule
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M.Tech from IIT Bombay(Expert maths Faculty)

Apply ILATE rules (standing for Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential) integration by parts rule!
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Math Tutor

firstly.....assuming 1+x=y, then integration by part e^y as 2nd function and 1/y as 1st ffunction
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M.Tech from IIT Bombay(Expert maths Faculty)

apply byparts rule... ILATE (standing for Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential)
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What are the different integration methods ?
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Dear friends, Please correct me in case I am wrong. I your opinion , what should be the maximum number of students in a coaching class, so that each student can have maximum dedicated time from the attending teacher in the classroom. In present days, cost and time of coaching / tuition matters a lot . The same will be demontrated by the following two examples : Example 1 : Suppose a student pays Rs. 4000/= per month ( 26 days in a month , excluding Sundays ), One hour per day , number of student is only one, then cost of coaching per minute works out to (Rs. 4000/=X 1 Student )/(26 days X 60 min/day ) = Rs.2.56 per min. cost and he will get dedicated time of 60 min. each day to carry on his studies. Example - 2 : Suppose a student pays Rs. 2000/= per month ( 26 days in a month , excluding Sundays ), One hour per day , number of student is 30 in a class, then cost of coaching per minute works out to (Rs. 2000/=X 30 Students in a coaching class ) / (26 days X 60 min/day ) = Rs.38.46 per min. cost and he will get dedicated time of only two min. each day for his studies. The above examples shows that though the monthly fees paid by each student is just half ( Rs. 2000/= see example -2 ) in case of classroom type of coaching as compared to the individual coaching ( Rs. 4000/= see example - 1 ), but on the other side cost of coaching per min. based on true dedicated time is 15 times higher in case of classroom type coaching with mass students than single student coaching. These can be however extrapolated for any number of students in a coaching class. As a result the speed of learning process of a student will be about fifteen times slower in classroom type coaching having mass students in a class ignoring the other factors. However, it may be the choice of the students and parents which option is better for them ? Regards, Sudhansu bhushan Roy.
Sir your calculations are excellent. I would like to add few things. 1. It depends on the student / parents choice at the end. 2. It never implies that effective attention to a student is 2-3 mins, so...
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What is the IUPAC Name of salicylic acid ?
2-hydroxy benzoic acid
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Dear students Differentiate x power underroot 2 with respect to x. Got rewarded by answering early.
first take y= x^(root 2) then taking log both the sides log y= (root 2) log x now diff w.r.to x 1/y (dy/dx)= (root 2) 1/x so dy/dx= y (root 2/x) now by putting y's value here dy/dx= x^(root 2)* (root 2/x)
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