How do you integrate #(x)/(x+10) dx#?

1 Answer
May 21, 2016

You could use two ways - pure algebra or partial fractions - either of which give you #x-10lnabs(x+10)+C# as the final answer.

Explanation:

Algebraic Approach

First, realize that we can rewrite the integral as:
#int(x+10-10)/(x+10)dx#

Now we can split it up into two fractions, like so:
#int(x+10)/(x+10)-10/(x+10)dx#
#=int1-10/(x+10)dx#

Using the sum rule for integrals, this further simplifies to:
#int1dx-int10/(x+10)dx#
#=int1dx-10int1/(x+10)dx#

Evaluating these is pretty straightforward now:
#x+C_1-10lnabs(x+10)+C_2#

Since #C_1+C_2# is just another constant, we can lump them together in one general constant #C#:
#intx/(x+10)dx = x-10lnabs(x+10)+C#

Partial Fractions Approach

Alternatively, if we want some practice with partial fractions or the teacher is forcing us to use this method, we can do it a little differently.

Since our original fraction #x/(x+10)# has only linear factors, we know the partial fraction decomposition will be something like:
#A+B/(x+10)#

Setting it up, we have:
#x/(x+10)=A+B/(x+10)#

Multiplying through by #x+10# gives us:
#x=A(x+10)+B#

If we let #x=-10#, we can find the value of #B#:
#x=A(x+10)+B#
#-10=A(-10+10)+B#
#-10=B#

Now we have:
#x=A(x+10)-10#

We can let #x# equal anything now to find #A#. For simplicity, let's have #x=0#:
#0=A(0+10)-10#
#10=10A#
#A=1#

Therefore #x/(x+10)=1-10/(x+10)#. Putting this back into the integral:
#int1-10/(x+10)dx#
#=x-10lnabs(x+10)+C-># as we discovered previously

I would not suggest the partial fractions method unless you were required to use it.