Author Topic: A123 Maths Bonanza. (Read 1253 times) Tweet Share

abd123 · « **on:** December 09, 2011, 09:27:57 am »

This question is a bit strange once, i got stuck on it yesterday night and tried to have a go again in the morning.

$\int_{-1}^{1}\frac{-1}{4-(x-1)^2}dx$
$I=\int_{-1}^{1}\frac{1}{(x-1)^2-4}dx$
$=\int_{-1}^{1}\frac{1}{(x-3)(x+1)}dx$

$=\frac{1}{4}\int_{-1}^{1}\frac{1}{x-3}-\frac{1}}{x+1}dx$
$=\frac{1}{4} [log_e\vert\frac{x-3}{x+1}\vert]_{-1}^{1}dx$
$=\frac{1}{4}(log_e(1)-log_e(\frac{4}{0})$

Helps Appreciated

dc302 · « **Reply #1 on:** December 09, 2011, 10:17:06 am »

Question must be wrong. What does the answer say?

ArchieMae · « **Reply #2 on:** December 09, 2011, 10:41:20 am »

Try doing the partial fractions without first multiplying by -1 as you did in step 1.

xZero · « **Reply #3 on:** December 09, 2011, 12:19:37 pm »

At x=-1, the graph has an asymptote so you can't integrate it at that point.

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Author Topic: A123 Maths Bonanza. (Read 1253 times) Tweet Share

abd123

A123 Maths Bonanza.

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ArchieMae

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xZero

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