Hi! I'm a second year Natural Science student at the University of Exeter. I love all things science, and enjoy sharing what I know about it with others.

For my GCSEs and A Levels, I found worked examples and practice questions are key. These helped me see where the material that I learned fit in with the rest of the subject, and I think the best test of understanding is to be able to use it on a problem. It's also great practice for exams! I will try and use this style in the lessons by first teaching the basic material, then guiding students through some examples. However, if this style doesn't work for some, I can easily change to suit the student.

Hi! I'm a second year Natural Science student at the University of Exeter. I love all things science, and enjoy sharing what I know about it with others.

For my GCSEs and A Levels, I found worked examples and practice questions are key. These helped me see where the material that I learned fit in with the rest of the subject, and I think the best test of understanding is to be able to use it on a problem. It's also great practice for exams! I will try and use this style in the lessons by first teaching the basic material, then guiding students through some examples. However, if this style doesn't work for some, I can easily change to suit the student.

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Johnny (Parent from Cambridge)

October 30 2016

Really appreciate you letting it run over and sorry about the internet problems.

This fraction can’t be integrated easily, but if we split it using partial fractions, these will be easier to integrate. To do this, we need to factoise the denominator, as this will follow the method of partial fractions:

^{2}∫_{1 }(6x+1) / (6x^{2}-7x+2) dx = ^{2}∫_{1 }(6x+1) / ((3x-2)(2x-1)) dx

We can then use partial fractions to split this fraction into two that can be integrated, by using the variables *A* and *B* to represent expressions that would multiply together to make 6x+1 when put above the parts of the factoised denominator.

^{2}∫_{1 }(6x+1) / ((3x-2)(2x-1)) dx = ^{2}∫_{1 }*A*/(3x-2) + *B*/(2x-1)dx

Following the method of partial fractions, cross-multiply the fractions:

^{2}∫_{1 }*A*/(3x-2) + *B*/(2x-1)dx = ^{2}∫_{1 }*A*(2x-1)/(3x-2) + *B*(3x-2)/(2x-1) dx

= ^{2}∫_{1 }(2*A*x – *A *+ 3*B*x – 2*B*)*/*((3x-2)(2x-1))* *dx

Therefore, we can say 2*A*x – *A *+ 3*B*x – 2*B = *6x+1, and so 2*A*x + 3*B*x = 6x and *-A *-2*B *= 1. We can then solve these simultaneous equations by elimination or substiution and find that *B* = -8 and *A* = 15. Therefore:

^{2}∫_{1 }(6x+1) / ((3x-2)(2x-1)) dx = ^{2}∫_{1 }*A*/(3x-2) + *B*/(2x-1) dx

= ^{2}∫_{1 }15/(3x-2) + -8/(2x-1) dx

= ^{2}∫_{1 }15/(3x-2) dx - ^{2}∫_{1} 8/(2x-1) dx

Using standard integrals, the integral of a fraction where the numerator is the derivative of the denominator is ln|denominator|. The fractions above are almost like this, if we rewrite them as:

^{2}∫_{1 }15/(3x-2) dx - ^{2}∫_{1} 8/(2x-1) dx = (5)^{2}∫_{1 }3/(3x-2) dx - (4)^{2}∫_{1} 2/(2x-1) dx

= 5[ln|3x-2|]^{2}_{1} – 4[ln|2x-1|]^{2}_{1}

= 5(ln(4) – ln(1)) – 4(ln(3) – ln(1))

= 5ln(4) – 4ln(3)

To get this into the form required for the question, we can use the law of logs: log(a^{b}) = b log(a):

5ln(4) – 4ln(3) = 5ln(2^{2}) – 4ln(3)

= 5(2)ln(2) – 4ln(3)

= 10ln(2) - 4ln(3)

This fraction can’t be integrated easily, but if we split it using partial fractions, these will be easier to integrate. To do this, we need to factoise the denominator, as this will follow the method of partial fractions:

^{2}∫_{1 }(6x+1) / (6x^{2}-7x+2) dx = ^{2}∫_{1 }(6x+1) / ((3x-2)(2x-1)) dx

We can then use partial fractions to split this fraction into two that can be integrated, by using the variables *A* and *B* to represent expressions that would multiply together to make 6x+1 when put above the parts of the factoised denominator.

^{2}∫_{1 }(6x+1) / ((3x-2)(2x-1)) dx = ^{2}∫_{1 }*A*/(3x-2) + *B*/(2x-1)dx

Following the method of partial fractions, cross-multiply the fractions:

^{2}∫_{1 }*A*/(3x-2) + *B*/(2x-1)dx = ^{2}∫_{1 }*A*(2x-1)/(3x-2) + *B*(3x-2)/(2x-1) dx

= ^{2}∫_{1 }(2*A*x – *A *+ 3*B*x – 2*B*)*/*((3x-2)(2x-1))* *dx

Therefore, we can say 2*A*x – *A *+ 3*B*x – 2*B = *6x+1, and so 2*A*x + 3*B*x = 6x and *-A *-2*B *= 1. We can then solve these simultaneous equations by elimination or substiution and find that *B* = -8 and *A* = 15. Therefore:

^{2}∫_{1 }(6x+1) / ((3x-2)(2x-1)) dx = ^{2}∫_{1 }*A*/(3x-2) + *B*/(2x-1) dx

= ^{2}∫_{1 }15/(3x-2) + -8/(2x-1) dx

= ^{2}∫_{1 }15/(3x-2) dx - ^{2}∫_{1} 8/(2x-1) dx

Using standard integrals, the integral of a fraction where the numerator is the derivative of the denominator is ln|denominator|. The fractions above are almost like this, if we rewrite them as:

^{2}∫_{1 }15/(3x-2) dx - ^{2}∫_{1} 8/(2x-1) dx = (5)^{2}∫_{1 }3/(3x-2) dx - (4)^{2}∫_{1} 2/(2x-1) dx

= 5[ln|3x-2|]^{2}_{1} – 4[ln|2x-1|]^{2}_{1}

= 5(ln(4) – ln(1)) – 4(ln(3) – ln(1))

= 5ln(4) – 4ln(3)

To get this into the form required for the question, we can use the law of logs: log(a^{b}) = b log(a):

5ln(4) – 4ln(3) = 5ln(2^{2}) – 4ln(3)

= 5(2)ln(2) – 4ln(3)

= 10ln(2) - 4ln(3)