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

$\dfrac{\log x}{2}\sqrt{16+(\log x)^{2}}+8\log\left|\log x+\sqrt{16+(\log x)^{2}}\right|+C$

CBSE Class 12 · 2005 · Indefinite IntegralsVery Short Answer

Evaluate :

\int \frac{16 + ( l o g x ) ^{2}}{x} d x

Answer: $\frac{lo g x}{2} 16 + (lo g x)^{2} + 8 lo g lo g x + 16 + (lo g x)^{2} + C$

Substitute

t = lo g x

, so

d t = \frac{1}{x} d x

. The integral turns into a standard

a^{2} + t^{2}

form with

a = 4

\int 16 + t^{2} d t = \frac{t}{2} 16 + t^{2} + \frac{16}{2} lo g t + 16 + t^{2} + C .

Putting back

t = lo g x

\frac{lo g x}{2} 16 + (lo g x)^{2} + 8 lo g lo g x + 16 + (lo g x)^{2} + C

CBSE Class 12 Board · 2005 · Subjective