Joseph W.

Currently unavailable: for new students

Degree: Chemical Physics (Masters) - Edinburgh University

Hey I'm Joe, I'm currently studying Chemistry and Physics at the University of Edinburgh. I'm a second year, and the pain of A levels still hasn't really left me! So I'd love to help you out and hopefully take a bit of the weight off your shoulders. I really enjoy explaining things and I've been told I'm quite good at it, so drop me a message and we'll go from there!

#### Subjects offered

SubjectLevelMy prices
Chemistry A Level £20 /hr
Maths A Level £20 /hr
Physics A Level £20 /hr
Chemistry GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr

#### Qualifications

MathematicsA-LevelA*
PhysicsA-LevelA*
ChemistryA-LevelA
Chemical Physics Year 1Bachelors Degree73%
 CRB/DBS Standard No CRB/DBS Enhanced No

#### Ratings and reviews

4.5from 2 customer reviews

Michelle (Parent) May 24 2016

Well paced. Clear explanations.

Cheng (Parent) May 18 2016

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### What is the integral of ln(x)? Hint: use parts for this integration

We use parts for this integration even though there is only one term! make ln(x)=1ln(x) How to do integration by parts If you have a function that you need to integrate that is two functions of x multiplied by eachother, and youve already tried everything else (inspection and substitution) t...

We use parts for this integration even though there is only one term! make ln(x)=1ln(x)

How to do integration by parts

If you have a function that you need to integrate that is two functions of x multiplied by eachother, and youve already tried everything else (inspection and substitution) then you're going to have to use integration by parts.

Intergrating the function §f(x)g(x) ...

we set one of the fuctions to u' and one to v

normally we'll take the hardest part to integrate as v.

but in this general example we'll just take f(x)=u' and g(x)=v (putting a ' on a fucntion is to say it has been differentiated)

The general equation then takes the form

§(u')v dx = uv - §(v')u dx

or §f(x)g(x) = (§f(x)dx)g(x) - §(§f(x)dx)g'(x)dx which just looks like a mess of symbols.

Sometimes youll have to do this more than once to get the answer, but its a very handy tool in integration!

So our problem is lnx. To make this by parts we split it up into two terms, 1(lnx)

Therefore u'=1 and v=lnx so u=x and v'=1/x

This makes the integral....

§-use this as integration sign

§ln(x)dx = xln(x) - §x(1/x)dx

=xln(x) -§(1)dx

= xln(x) -x + c

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1 year ago

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