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C3 help urgent!

Can some1 please show me how to do question 4b.
http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MPC3-W-QP-JAN07.PDF

I can do simple substitution but not the harder ones :frown:.
If you could show me all the steps that you did to get the answer, that would be great.
Ty for reading :colondollar:
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Avatar for raheem94
raheem94
13
Original post by bahjat93
Can some1 please show me how to do question 4b.
http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MPC3-W-QP-JAN07.PDF

I can do simple substitution but not the harder ones :frown:.
If you could show me all the steps that you did to get the answer, that would be great.
Ty for reading :colondollar:


It isn't a harder one!

∫xx2+5 dx   u=x2+5 \displaystyle \int x \sqrt{x^2+5} \ dx \ \ \ u = x^2 +5

Differentiate, u=x2+5 u = x^2 +5
This gives, dx=du2x \displaystyle dx = \frac{du}{2x}

The integral becomes, ∫xx2+5 dx=∫xu12×du2x=∫u122 du \displaystyle \int x \sqrt{x^2+5} \ dx = \int x u^{\frac12} \times \frac{du}{2x} = \int \frac{u^{\frac12}}{2} \ du

Now its simple to finish it off.
Reply 2
Avatar for bahjat93
bahjat93
OP
19
Original post by raheem94
It isn't a harder one!

∫xx2+5 dx   u=x2+5 \displaystyle \int x \sqrt{x^2+5} \ dx \ \ \ u = x^2 +5

Differentiate, u=x2+5 u = x^2 +5
This gives, dx=du2x \displaystyle dx = \frac{du}{2x}

The integral becomes, ∫xx2+5 dx=∫xu12×du2x=∫u122 du \displaystyle \int x \sqrt{x^2+5} \ dx = \int x u^{\frac12} \times \frac{du}{2x} = \int \frac{u^{\frac12}}{2} \ du

Now its simple to finish it off.


I don't get maths but i see what you did, LOOOL cheers dude much appreciated

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