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So I'm to find the definite integral of a function which I'm to convert into partial fractions.

$$\int_0^1 \frac{2}{2x^2+3x+1}\,dx$$

Converting to partial fractions I get... $\frac{A}{2x+1} + \frac{B}{x+1}$ with $A = 4$ and $B = -2$

Thus the definite integral is...

$$ \begin{align} & \int_0^1 \left(\frac{4}{2x+1}-\frac{2}{x+1}\right)\,dx \\[8pt] & =[4\ln|2x+1| - 2\ln|x+1|]_0^1 \\[8pt] & = 4\ln|3|-2\ln|2| - (4\ln|1| - 2\ln|1|) \\[8pt] & = 4\ln|3| - 2\ln|2| - 0 \\[8pt] & = 2(2\ln|3|-\ln|2|) \\[8pt] & = 2\ln\left|\frac{9}{2}\right| \end{align} $$

However, the answer in the book gives $2\ln|\frac{3}{2}|$ as do online integral calculators, so I imagine I've done something wrong, but can't for the life of we work out what since I keep getting the same values for A and B and the same answer.

Any ideas?

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1 Answer 1

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You neglected the chain rule: $$ \int \frac{4}{2x+1} \, dx = 2\ln|2x+1|+C \ne 4\ln|2x+1|+C. $$

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  • $\begingroup$ Thank you! That's what I did wrong. $\endgroup$
    – Wolff
    Commented Mar 2, 2014 at 16:49

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