6.3. Differentiation and Integration of Logarithmic and Exponential Functions http://www.ck12.org
∫ 1
xdx=ln|x|+C.Example 6:
Evaluate∫x+^11 dx.
Solution:
In general, whenever you encounter an integral with an integrand as a rational function, it might be possible that it
can be integrated with the rule of natural logarithm. To do so, determine the derivative of the denominator. If it is
the numerator itself, then the integration is simply the ln of the absolute value of the denominator. Let’s test this
technique.
∫ 1
x+ 1 dx.Notice that the derivative of the denominator is 1, which is equal to the numerator. Thus the solution is simply the
natural logarithm of the absolute value of the denominator:
∫ 1
x+ 1 dx=ln|x+^1 |+C.The formal way of solving such integrals is to useu−substitution by lettinguequal the denominator. Here, let
u=x+ 1 ,anddu=dx.Substituting,
∫ 1
x+ 1 dx=∫ 1
udu
=ln|u|+C
=ln|x+ 1 |+C.Remark:The integral must use the absolute value symbol because althoughxmay have negative values, the domain
of ln(x)is restricted tox≥ 0.
Example 7:
Evaluate∫ 4 x (^24) +x+ 2 x^1 + 1 dx.
Solution:
As you can see here, the derivative of the denominator is 8x+ 2 .Our numerator is 4x+ 1 .However, when we multiply
the numerator by 2,we get the derivative of the denominator. Hence
∫ 4 x+ 1
4 x^2 + 2 x+ 1 dx=
1
2
∫ 2 ( 4 x+ 1 )
4 x^2 + 2 x+ 1 dx
=^12∫ 8 x+ 2
4 x^2 + 2 x+ 1 dx
=^12 ln| 4 x^2 + 2 x+ 1 |+C.Again, we could have usedu−substitution.