# Math Help - integral

1. ## integral

int((1/x)^3)/x

(1/x^3)/x

(1/x^3)*(x^-1)

x^-3*x^-1

x^4

(1/5)x^5

2. Originally Posted by xcelxp

int((1/x)^3)/x

(1/x^3)/x

(1/x^3)*(x^-1)

x^-3*x^-1

x^4

(1/5)x^5
i have no idea what you are trying to do. why did you change ln(x) to 1/x? here:

3. ## some additional helps for you

See, I want to say you some generalised idea about how to integrate such kind of composite functions where you have two kind of functions. In this question you have one logarithmic function and another algebraic function(1/x). In this kind of problem you do use a formula called ILATE: I is for Inverse, L for logarithm, A for algebraic, T for trigonometric and E for exponential functions. Just choose the first function depending upon the sequence from the start. Then move on to integrate as following

Suppose consider the integral as I1 and then you can write

I1=first function * (integration of 2nd function)- integration of ((derivative of the first function)*(integration of second function))

This way you can solve most of the questions like this.

If you want any further help you may send e-mail to me @gmail.com

Good luck

4. Originally Posted by jagabandhu
See, I want to say you some generalised idea about how to integrate such kind of composite functions where you have two kind of functions. In this question you have one logarithmic function and another algebraic function(1/x). In this kind of problem you do use a formula called ILATE: I is for Inverse, L for logarithm, A for algebraic, T for trigonometric and E for exponential functions. Just choose the first function depending upon the sequence from the start. Then move on to integrate as following

Suppose consider the integral as I1 and then you can write

I1=first function * (integration of 2nd function)- integration of ((derivative of the first function)*(integration of second function))

This way you can solve most of the questions like this.

If you want any further help you may send e-mail to me @gmail.com

Good luck
what you have said is all well and good, but what you are describing is the Integration By Parts formula. It will work in this instance and many others like it, however, i feel it is an overkill for a problem of this nature. Substitution is usually the easier of the two methods, if its applicable, at least in my experience