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

Question

Ezra Adams · Accepted Answer

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

Algebraic Methodology 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# Approach Using Partial Fractions Alternatively, we can go about things a little differently if we'd like to practice with partial fractions or if the teacher is making us use this approach. 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 If you weren't required to use the partial fractions method, I wouldn't recommend it.

Emilia Aguilar · Answer

undefined To integrate (x)/(x+10) dx, you can use the method of partial fractions. After decomposition, the integral becomes the sum of two simpler integrals:

∫(x)/(x+10) dx = ∫(10/(x+10) - 10/(x+10)^2) dx

Integrate each term separately:

= 10 ln|x+10| - 10/(x+10) + C

So, the integral of (x)/(x+10) dx is 10 ln|x+10| - 10/(x+10) + C, where C is the constant of integration.