I was wondering how you integrate - with simply the integral of the power rule (n^x)' = nx^n-1, when the equation, such as (15x^3 + 8x^2 - 5/x) doesn't have any multiples for the antiderivative of 15x^3 or 8x^2...

 

do you just leave them or something...

I do not understand what you are asking.

 

Integration is linear so you can integrate each term one by one. The only trouble you may have is that using the simple "power rule" as you put it will not help you integrate 1/x, at least not very directly.

sorry, to be clearer, i meant, for a term like 15x^3 or 8x^2, how do you integrate them when the only way - to my knowledge - is using the antiderivative of the power rule, so adding 1 to the power instead. As there aren't any multiples of 4 or 3 which give you the bases (15 & 8)...

Example:

 

[math] \int dx \: 15 x^{3} = \frac{15}{4}x^{4} + Const[/math].

so in the situation of no multiples of the reversed power (x^n+1 instead of x^n-1 for the derivative) to equal the base, you just leave the base and do it to the rest of the term..?

so in the situation of no multiples of the reversed power (x^n+1 instead of x^n-1 for the derivative) to equal the base, you just leave the base and do it to the rest of the term..?

 

You just need to divide by the right number to "reverse engineer" your answer.

yes sorry, i understand now. i was looking at it upside down...

 

cheers.

For simple polynomials and standard elementary functions ask yourself what function do I differentiate to get the function I want to integrate?

 

Have a guess and then try differentiating it and see if you are right. Most likely the first few examples you try you will be out by simple numerical factors, or a sign when trig functions are used. You can then "fudge" it to get the correct answer.

 

This only works with things you can spot easily and this improves with practice.

sorry, one more thing, with u substitution, why exactly do you bother in the processes of differentiating u when the end product requires the integral form ?

you let [math]u[/math] be some function of [math]x[/math]. The initial integral is with respect to [math]dx[/math]. Thus you need to change the variables to get at an integral wrt [math]du[/math], hopefully this integral will be easier to evaluate.

 

Let consider a very made up example.

 

[math]I = \int dx\: 2 x(x^{2}+1)^{2}[/math].

 

Let us use the substitution [math]u = x^{2}+1[/math], then

 

[math]\frac{du}{dx} = 2x[/math],

 

thus

 

[math]du = 2x \: dx[/math] .

 

Then we can write

 

[math]I = \int du \: u^{2} = \frac{1}{3}u^{3}+ c[/math].

 

Undoing the substitution gives

 

[math]I = \frac{1}{3}(x^{2}+1)^{3}+ c[/math],

 

which we can write as

 

[math]I = \frac{1}{3}x^{6} + x^{4} + x^{2} + k[/math].

 

To recap, you need to change the integrate from [math]dx[/math] to [math]du[/math]. To do this differentiate [math]u[/math]. Simplify the integral now in terms of [math]u[/math]. Evaluate the integral and then undo the substitution.

 

(Corrected typo as pointed out by Shadow)

Edited September 17, 201114 yr by ajb

thus

 

[math]du = 2x \: du[/math] .

 

Just to avoid possible confusion, this is supposed to be [math]du = 2x \: dx[/math].

Edited September 17, 201114 yr by Shadow

Just to avoid possible confusion, this is supposed to be [math]du = 2x \: dx[/math].

 

Thank for that, it has now been corrected.

also, it may seem basic, but what is the derivative of x/2 ?

also, it may seem basic, but what is the derivative of x/2 ?

 

1/2

also, it may seem basic, but what is the derivative of x/2 ?

 

If you are not sure "act stupid" and apply the rules of derivatives.

 

[math]\frac{d}{dx} \left( \frac{1}{2} \times x \right) = \frac{d}{dx}\left(\frac{1}{2}\right)\times x + \frac{1}{2} \times \frac{d}{dx} x[/math].

 

Now the half is a constant as far as x is concerned, thus the derivative vanishes. The rate of change of x with respect to x is just one. So

 

[math]\frac{d}{dx} \left( \frac{1}{2} \times x \right) = \frac{1}{2} \times \frac{dx}{dx} = \frac{1}{2}[/math].