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.

3 Answers
Jun 6, 2018

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.

Jun 6, 2018

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)

Jun 6, 2018

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#