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Integrating e^x & 1/x (Cambridge (CIE) O Level Additional Maths) : Revision Note

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Integrating e^x & 1/x

How do I integrate exponentials and 1/x?

  • The integrals involvingbold space bold e to the power of bold italic x andspace bold ln bold space bold italic x are

bold space bold integral bold space bold e to the power of bold italic x bold space bold d bold italic x bold equals bold space bold e to the power of bold italic x bold plus bold italic c

where bold italic c is the constant of integration

  • For the linear functionbold space bold left parenthesis bold italic a bold italic x bold plus bold italic b bold right parenthesis, wherespace bold italic a andspace bold italic b are constants,

 bold space bold integral bold e to the power of bold italic a bold italic x bold plus bold italic b end exponent bold space bold d bold italic x bold equals bold 1 over bold italic a bold e to the power of bold italic a bold italic x bold plus bold italic b end exponent bold plus bold italic c

  • It follows from the last result that

 

  • which can be deduced using Reverse Chain Rule

  • With ln, it can be useful to write the constant of integration,space c, as a logarithm

    • using the laws of logarithms, the answer can be written as a single term

    • wherespace k is a constant

    • This is similar to the special case of differentiatingspace ln space left parenthesis a x plus b right parenthesis whenspace b equals 0

Worked Example

A curve has the gradient functionspace f apostrophe left parenthesis x right parenthesis equals fraction numerator 3 over denominator 3 x plus 2 end fraction plus straight e to the power of 4 minus x end exponent.

Given the exact value ofspace f left parenthesis 1 right parenthesis isspace ln space 10 minus straight e cubed find an expression forspace f left parenthesis x right parenthesis.

 

To find f open parentheses x close parentheses, we need to integrate the expression for f apostrophe open parentheses x close parentheses.   

f open parentheses x close parentheses equals integral fraction numerator 3 over denominator 3 x plus 2 end fraction plus e to the power of 4 minus x end exponent space d x  

Rewrite as two separate integrals which can be found individually, and factor out any constants.  

f open parentheses x close parentheses equals 3 integral fraction numerator 1 over denominator 3 x plus 2 end fraction d x space plus space integral e to the power of 4 minus x end exponent space d x  

Use the two results: integral fraction numerator 1 over denominator a x plus b end fraction d x equals 1 over a ln vertical line a x plus b vertical line space plus space c integral e to the power of a x plus b end exponent d x space equals fraction numerator space 1 over denominator a end fraction space e to the power of a x plus b end exponent space plus c   

f open parentheses x close parentheses equals 3 open parentheses 1 third ln vertical line 3 x plus 2 vertical line close parentheses space plus space minus 1 e to the power of 4 minus x end exponent space plus c f open parentheses x close parentheses equals ln vertical line 3 x plus 2 vertical line space minus space e to the power of 4 minus x end exponent space plus c  

The question states that f open parentheses 1 close parentheses equals ln 10 space minus space e cubed, so substitute in x equals 1 and equate to the given expression.  

ln space 10 space minus space e cubed space equals space ln vertical line 3 open parentheses 1 close parentheses plus 2 vertical line space minus space e to the power of 4 minus 1 end exponent plus c  

Simplify and solve to find c.  

ln space 10 space minus space e cubed space equals space ln space 5 space minus space e cubed space plus space c ln space 10 minus ln space 5 space equals space c  

Recall from laws of logarithms that ln space a space minus space ln space b space equals ln space a over b space.  

c equals ln space 10 over 5 equals ln space 2  

Rewrite the full expression for f open parentheses x close parentheses.  

 
or by combining logs, this could be written as f open parentheses x close parentheses equals ln space open parentheses 2 vertical line 3 x plus 2 vertical line close parentheses space minus space e to the power of 4 minus x end exponent  

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    Paul

    Author: Paul

    Expertise: Maths Content Creator (Previous)

    Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams.