Question

Evaluate: $\int \frac{ \sqrt{1+(\log x)^{2}}}{x} \ dx$

Answer 1

solution:

This integration can be done by substitution Method.

Let

log x = t

1/x dx = dt

Substitute this value in main equation

$\sqrt{1+t^{2}}\: dt$

This is the standard form of Integration,which can integrate as

$\int\sqrt{a^{2}+x^{2}} dx= \frac{x}{2}\sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2}log(x+\sqrt{a^{2}+x^{2}}$

so here a=1

$\int\sqrt{1^{2}+t^{2}} dt= \frac{t}{2}\sqrt{1^{2}+t^{2}}+\frac{1^{2}}{2}log(t+\sqrt{1^{2}+t^{2}}$

Redo substitution

$=\frac{log\:x}{2}\sqrt{1^{2}+(log\:x)^{2}}+\frac{1^{2}}{2}log(log x+\sqrt{1^{2}+(log x)^{2}}+C\\\\=\frac{log\:x}{2}\sqrt{1+(log\:x)^{2}}+\frac{1}{2}log(log x+\sqrt{1+(log x)^{2}}+C$