Integration Problem

**TaylorM0192** · July 26th 2010, 07:34 PM

I'm integrating $\frac{sqrt(x+4)}{x}$ . After doing a substitution I get to $\frac{u^2}{u^2-4}$ .

I want to integrate this with partial fractions, but after going through the process I get $\frac{1}{u-2} - \frac{1}{u+2}$ which is evidently not the correct solution. When I finished integrating the function I got to the correct answer, except for one missing term. The missing term indicates the partial fraction decomposition is (and I confirmed it) $1+\frac{1}{u-2} - \frac{1}{u+2}$ .

I'm not asking for anyone in integrate the function - I just want to know why the method of partial fractions fails in this instance?

Heh, just realized my mistake (the rational function not being in proper form for partial fraction decomposition). Sorry for the post; I solved the problem.

**Prove It** · July 26th 2010, 11:23 PM

Partial Fractions doesn't fail...

You have $\int{\frac{\sqrt{x + 4}}{x}\,dx} = \int{\frac{(\sqrt{x + 4})^2}{x\sqrt{x + 4}}\,dx}$

$= \int{\frac{2(\sqrt{x + 4})^2}{x}\cdot \frac{1}{2\sqrt{x + 4}}\,dx}$ .

Now let $u = \sqrt{x + 4}$ so that $du = \frac{1}{2\sqrt{x + 4}}\,dx$ , and note that $x = u^2 - 4$ , and the integral becomes

$\int{\frac{2u^2}{u^2 - 4}\,du}$

$= 2\int{1 + \frac{4}{u^2 - 4}\,du}$ after long division

Now applying Partial Fractions...

$\frac{A}{u + 2} + \frac{B}{u - 2} = \frac{4}{u^2 - 4}$

$\frac{A(u - 2) + B(u + 2)}{u^2 - 4} = \frac{4}{u^2 - 4}$

$A(u - 2) + B(u + 2) = 4$

$Au - 2A + Bu + 2B = 4$

$(A + B)u - 2A + 2B = 0u + 4$ .

Therefore $A + B = 0$ and $-2A + 2B = 4$ .

Solving these simultaneously gives $A = -1$ and $B = 1$ .

Now so the integral becomes

$2\int{1 - \frac{1}{u + 2} + \frac{1}{u - 2}\,du}$ .

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