# Evaluate :$\int\large\frac{\sqrt{16+(\log x)^2}}{x}$$dx ## 1 Answer Toolbox: • Method of substitution: • Given \int f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t). • Consider I=\int f(x)dx. • Put x=g(t) so that \frac{dx}{dx}=g'(t). • dx=g'(t)dt. • Thus I=\int f(g(t).g'(t))dt. • \int \sqrt{a^2+x^2}dx=\large\frac{x}{2}$$\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,
$\large\frac{1}{x}$$dx=dt \therefore I=\int\sqrt{16+t^2}dt This is of the form \int \sqrt{a^2+x^2}dx=\large\frac{x}{2}$$\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,