SOLUTION: I have 2 indefinite integral problems I am struggling with, please help me if you can. #1. dx/3x^10=?+c #2. (3+4u)/u du =?

Algebra ->  Equations -> SOLUTION: I have 2 indefinite integral problems I am struggling with, please help me if you can. #1. dx/3x^10=?+c #2. (3+4u)/u du =?      Log On


   

Question 534008: I have 2 indefinite integral problems I am struggling with, please help me if you can.
#1. dx/3x^10=?+c #2. (3+4u)/u du =?
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
First problem: int%281%2F%283%2Ax%5E10%29%2Cdx%29+
.
Bring the constant 1%2F3 outside the integration sign.
.
%281%2F3%29%2Aint%28%281%2Fx%5E10%29%2Cdx%29+
.
Recall that if a variable in the denominator has a positive exponent, it can be brought into the numerator with a negative exponent. So, the x%5E10 in the denominator can be moved into the numerator as x%5E%28-10%29 changing the problem to:
.
%281%2F3%29%2Aint%28%28x%5E%28-10%29%29%2Cdx%29+
.
Now integrate using the rule int%28u%5En%2Cdx%29+=+u%5E%28n%2B1%29%2F%28n%2B1%29
.
This rule results in the integration becoming:
.

.
Simplify the answer by adding the -10+1 in two places to get the answer:
.
%281%2F3%29%2A%28x%5E%28-9%29%2F%28-9%29%29
.
Further simplify the answer by factoring out the -9 from the denominator and multiplying it times the 3 in the denominator of the multiplier 1%2F3 to make it:
.
%28-1%2F27%29%2A%28x%5E%28-9%29%29
.
It would make things a little more conventional to move the variable x into the denominator by changing its exponent from -9 to +9 in doing so:
.
%28%28-1%29%2F27%29%2A%281%2Fx%5E9%29
.
Finally, since this was an indefinite integration, don't forget to include the constant c in the answer ... thereby making the final answer:
.
%28%28-1%29%2F27%29%2A%281%2Fx%5E9%29%2Bc
.
You could also write this as:
.
%28%28-1%29%2F%2827x%5E9%29%29%2B+c
.
If you so choose to do so.
.
Second problem:
.
int%28%283%2B4u%29%2Fu%2Cdu%29
.
Begin splitting this into two separate problems by dividing the u into the denominator into both terms of the numerator as follows:
.
int%28%28%283%2Fu%29%2B%284u%2Fu%29%29%2Cdu%29
.
Now multiply both of these two terms by du and you change the problem to:
.
int%28%283%2Fu%29%2Cdu%29%2Bint%28%284u%2Fu%29%2Cdu%29
.
In the first integration move the constant 3 outside the integration sign and in the second integration move the constant 4 outside the integration sign. When you do those two moves you have:
.
3%2Aint%28%281%2Fu%29%2Cdu%29%2B4%2Aint%28%28u%2Fu%29%2Cdu%29
.
You should recognize the first integration is of the standard form
.
int%281%2Fx%2Cdx%29+=+ln%28abs%28x%29%29
.
Note that the x is an absolute value.
.
Completing the first integration and multiplying it by the constant 3 results in the problem being reduced to:
.
3%2Aln%28abs%28u%29%29+%2B+4%2Aint%28%28u%2Fu%29%2Cdu%29
.
In the second integral notice that u%2Fu+=+1 so you are left with the following:
.
3%2Aln%28abs%28u%29%29+%2B+4%2Aint%281%2Cdu%29
.
and the integration of du results in just u. Since this is problem involves just indefinite integrals the constant c must be added and the result becomes:
.
3%2Aln%28abs%28u%29%29+%2B+4%2Au+%2B+c
.
That's it. Hope that these two problems help you to understand better the process of integration or anti-derivation, whatever you have been taught to call it.
.