Question

Are there multiple answers to integrals?

`int(1/(6x))dx`
Which integrates to:
`1/6ln(x) +c`
However, I also attempted to use another method of integrating and retrieved:
`1/6ln(6x)+c`

I am confused as they both derive the same-original equation yet are different answers. Is anyone able to help me understand why this is.
Thanks in advance.

Answer 1

No, an integral will only give one solution (or in general, one class of solutions that differ by a constant term)

Explanation:

The first solution `1/6 ln(x) + c` is the correct solution.

I'm not sure how you arrived at the second answer, but there must have been a mistake with your computation.

Some times there are equivalent forms of solutions for integrals, but in this case the `6` inside of the argument of the `ln` is not equivalent. Usually this only occurs when you have trigonometric functions, as they can be expressed as other trig functions or exponentials.

Answer 2

Both of the the answer are consistent:

`1/6ln(6x)+c = 1/6ln6 + 1/6lnx + c`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1/6lnx + A` where `A=c+1/6ln6`

Differentiation is a unique 1:1 function transformation, By FTC Integration is therefore unique but can differ by a constant (as the derivative of a constant is zero)

Answer 3

There are more than one answer in different form is
possible ,but not the different answer .

Here the different form means external look.!!

.But mathematically both the answers are same .

Explanation:

Before I answer see the both methods.

`(i)int(1/(6x))dx=1/6int1/xdx=color(red)(1/6ln|x|+c`

`(ii)int(1/(6x))dx`.

Let , `6x=u=>6dx=du=>dx=1/6du`

`:.int(1/(6x))dx=int(1/u)*1/6du=1/6ln|u|+c=color(red)(1/6ln|6x|+c`

So, your second answer is 100% correct.There is no mistake
with your computation.

There are more than one answer in different form is
possible ,but not the different answer .
Here the different form means external look.!!

.But mathematically both the answers are same .

Also, The subtraction of both the answers is always CONSTANT.

If we take, `(i)f(x)=color(red)(1/6ln|x|) and (ii)g(x)=color(red)(1/6ln|6x|) ,then`

`g(x)-f(x)=1/6ln|6x|-1/6ln|x|=1/6{ln|6x|-ln|x|}`

`:.g(x)-f(x)=1/6{ln|6|+ln|x|-ln|x|}=1/6ln6`

`i.e. color(blue)( g(x)-f(x)=CONSTANT`

Please see the illustrations carefully.

`(1)int1/(x^2+a^2)dx=1/atan^-1(x/a)+c=-1/acot^-1(x/a)+c`

`(2)intsecxdx=ln|tan(pi/4+x/2)|+c=ln|secx+tanx|+c`

`(3)intcsc(2x)dx=1/2ln|csc2x-cot2x|+c=1/2ln|tanx|+c`