I'm trying to integrate: $$\int \frac{dx}{x(7+\log^{2}x)}$$
Can it be done without trig substitutions or without using $\tan^{-1}$?
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$\begingroup$ It is done by substituting $t=lnx$ and not trigonometric substitution . But answer comes in terms of $tan^{-1}$ $\endgroup$ – ABC Mar 31 '13 at 17:09
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$\begingroup$ can i bypass the $tan^{-1}$ term? because i don't know how to get it after putting $t=lnx$ $\endgroup$ – user64370 Mar 31 '13 at 17:12
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3$\begingroup$ Avoid double dollar signs on titles. $\endgroup$ – Pedro Tamaroff♦ Mar 31 '13 at 17:14
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$\begingroup$ @PeterTamaroff - ok. $\endgroup$ – user64370 Mar 31 '13 at 17:17
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The final answer involves $\arctan$ and hence you cannot avoid trigonometric substitution.
Let $\ln(x) = t$. We then have $dt = \dfrac{dx}x$. Hence, we have $$I = \int \dfrac{dx}{x(7 + \ln^2(x))} = \int \dfrac{dt}{7+t^2}$$ Now set $t = \sqrt7 \tan(y)$. We then have $$I = \int \dfrac{\sqrt7 \sec^2(y) dy}{7 \sec^2(y)} = \dfrac1{\sqrt7} \arctan \left(\dfrac{t}{\sqrt{7}}\right) + \text{constant} = \dfrac1{\sqrt7} \arctan \left(\dfrac{\ln(x)}{\sqrt{7}}\right) + \text{constant}$$
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Hint: for every differentiable function $u$, $$ \int\frac{u'(x)\mathrm dx}{7+u(x)^2}\stackrel{t=u(x)}{=}\int\frac{\mathrm dt}{7+t^2}\stackrel{t=\sqrt{7}s}{=}\frac1{\sqrt{7}}\int\frac{\mathrm ds}{1+s^2}=\text{____}. $$
