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Evaluate :$\int\large\frac{\sqrt{16+(\log x)^2}}{x}$$dx$

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  • Method of substitution:
  • Given $\int f(x)dx$ can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dx}=g'(t).$
  • dx=g'(t)dt.
  • Thus $ I=\int f(g(t).g'(t))dt.$
  • $\int \sqrt{a^2+x^2}dx=\large\frac{x}{2}$$\sqrt{x^2+a^2}+\large\frac{a^2}{2}$$\log \mid x+\sqrt{x^2+a^2}\mid+c$
Step 1:
Let $I=\int \large\frac{16+(\log x)^2}{x}$$dx$
Put $\log x=t$
Differentiating with respect to $x$ we get,
$\therefore I=\int\sqrt{16+t^2}dt$
This is of the form
$\int \sqrt{a^2+x^2}dx=\large\frac{x}{2}$$\sqrt{x^2+a^2}+\large\frac{a^2}{2}$$\log \mid x+\sqrt{x^2+a^2}\mid+c$
Step 2:
Here $a=4$ and $x=t$
$\therefore \sqrt{16+t^2}dt=\large\frac{t}{2}$$\sqrt{t^2+4^2}+\large\frac{4^2}{2}$$\log\mid t+\sqrt{t^2+16}\mid+c$
$\Rightarrow \large\frac{1}{2}$$\sqrt{t^2+16}+8\log \mid t+\sqrt{t^2+16}\mid +c$
Step 3:
Substituting for $t$ we get,
$\large\frac{\log x}{2}$$\sqrt{(\log x)^2+16}+8\log\mid \log x+\sqrt{(\log x)^2+16}\mid +c$
answered Sep 20, 2013 by sreemathi.v
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