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Integrals of Exponential and Logarithmic Functions

Function Integral
lnx xlnx - x + c
logx (xlnx - x) / ln(10) + c
logax x(logax - logae) + c
ex ex+c
ek∙x 1 / k ∙ ek∙x + c
ax ax / lna + c
xn 1 / (n+1) ∙ xn+1 + c, where |n|≠ 1 
1/x = x-1 ln|x|+c
x = x1/2 2/3 ∙ (√x)3 + c = 2/3 ∙ x3/2 + c, where c is a constant

Example 1: Solve integral of exponential function ∫ex32x3dx

Solution

Step 1: the given function is ∫ex^33x2dx

Step 2: Let u = x3 and du = 3x2dx

Step 3: Now we have: ∫ex^33x2dx= ∫eudu

Step 4: According to the properties listed above: ∫exdx = ex+c, therefore ∫eudu = eu + c

Step 5: Since u = x3 we now have ∫eudu = ∫ex3dx = ex^3 + c

So the answer is ex^3 + c



Example 2:  Integrate .

Solution: First, split the function into two parts, so that we get:




Example 3: Integrate lnx dx.

Solution: 

Let

u=lnu

and

dv = dx = (1)dx

 

so that:

du = 1/x dx

and

v = x

 

Therefore:

    lnx dx   

= x lnx - ∫x * 1/x dx  

=xlnx - ∫1dx  

=xlnx – x + C



Example 4: Integrate .

Solution:

Use u-substitution. Let u = 3 + lnx

So that du = 1/x dx.

Substitute into the original problem we will get:




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