definite integral (1 Viewer)

juantheron · Jun 23, 2012

$\displaystyle \bf{\int_{1}^{3}\frac{\ln(2x+2)}{x^2+2x+15}dx}$

$\bf{My \;\; Solution::}$

$\bf{Let \;\; (x+1) = t\Leftrightarrow dx = dt\;\; and \;\; changing \;\; Limit\;,\; We\;\; Get}$

$\bf{\int_{2}^{4}\frac{\ln(2t)}{t^2+\left(\sqrt{14}\right)^2}dx }$

$\bf{Using\;\; I.B.P\;,We\;\; Get}$

$\bf{\left[\ln(2t).\frac{1}{\sqrt{14}}.\tan^{-1}\left(\frac{t}{\sqrt{14}}\right)\right]_{2}^{4}-\int_{2}^{4}\frac{1}{2t}.\frac{1}{\sqrt{14}}.\tan^{-1}\left(\frac{t}{\sqrt{14}}\right)dt}$

Now my Question is How can I solve The (2) Integral means

$\bf{\int \frac{1}{u}.\tan^{-1}(u)}du$ Type

Thanks

Carrotsticks · Jun 23, 2012

Use Integration by parts with the choices u=arctan(x) and dv=1/x

Then upon doing so, use Partial fractions for the integral of:

$\frac{1}{x^2(1+x^2)}$

Which decomposes to:

$\frac{1}{x^2(1+x^2)} = \frac{1}{x^2} - \frac{1}{1+x^2}$

definite integral (1 Viewer)

juantheron

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Carrotsticks

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