Help with integration!

**Tutu** · September 22nd 2013, 09:48 AM

Hi

I really need help with these two questions, for the first attachement I have no idea how to do that last part circled in pink

For the second attachement(labeled part e.)), I could not get the answer - I got $\frac{4sqrt{2}}{sqrt{3}}arctan(\frac{sqrt{2}}{sqrt {3}}) + c$ but the answer is $\2\frac{sqrt{2}}{sqrt{3}}arctan(\frac{sqrt{2}}{sqr t{3}}) + c$

Thank you so much in advance!

**FelixFelicis28** · September 22nd 2013, 10:29 AM

Originally Posted by Tutu

Hi

I really need help with these two questions, for the first attachement I have no idea how to do that last part circled in pink

For the second attachement(labeled part e.)), I could not get the answer - I got $\frac{4sqrt{2}}{sqrt{3}}arctan(\frac{sqrt{2}}{sqrt {3}}) + c$ but the answer is $\2\frac{sqrt{2}}{sqrt{3}}arctan(\frac{sqrt{2}}{sqr t{3}}) + c$

Thank you so much in advance!

1) You have $\int_0^3 f(x) \ dx = 8 = \int_c^d f(x-2) \ dx$

Integration by substitution - the substitution should be pretty easy to spot, bearing in mind the argument of $f(x-2)$ .

2) Their answer is correct, I cannot point out what you've done incorrectly unless you post your working OP, likely a small arithmetical error somewhere.

**Tutu** · September 22nd 2013, 07:58 PM

Hi thank you so much!

For the second, attached is my working, I had simply used the formula so its sort of one step..can you help me see where I went wrong?

**FelixFelicis28** · September 22nd 2013, 08:10 PM

Originally Posted by Tutu

Hi thank you so much!

For the second, attached is my working, I had simply used the formula so its sort of one step..can you help me see where I went wrong?

You haven't done $\int \frac{dx}{a^2 + (px+q)^2} = \frac{1}{ap} \arctan \left(\frac{px+q}{a}\right) + \mathcal{C}$

but rather:

$\int \frac{dx}{a^2 + (px+q)^2} = \frac{1}{a} \arctan \left(\frac{px+q}{a}\right) + \mathcal{C}$ .

**ibdutt** · September 22nd 2013, 08:54 PM

Math Help - Help with integration!

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