Question

$\int_{}^{}{\frac{dx}{x\log x\log(\log x)} =}$

• A. $2\log(\log x) + c$
• B. $\log\lbrack\log(\log x)\rbrack + c$
• C. $\log(x\log x) + c$
• D. $\int_{}^{}{\frac{\sec^{2}x\mspace{6mu} dx}{\sqrt{\tan^{2}x + 4}} =}$

Answer 1

Answer: (B)

$\log\lbrack\log(\log x)\rbrack + c$

Answer 2

$\cos^{- 1}a^{x} + c$

Multiplying $\int_{}^{}\frac{\sqrt{\tan x}}{\sin x\cos x}\mspace{6mu} dx =$ and $2\sqrt{\sec x} + c$ by $2\sqrt{\tan x} + c$ we get

$\frac{2}{\sqrt{\tan x}} + c$

$\frac{2}{\sqrt{\sec x}} + c$

$\int_{}^{}\frac{\sin 2x}{a^{2} + b^{2}\sin^{2}x}\mspace{6mu} dx =$

$\frac{1}{b^{2}}\log(a^{2} + b^{2}\sin^{2}x) + c$.

Trick : By inspection,

$\frac{1}{b}\log(a^{2} + b^{2}\sin^{2}x) + c$

$\log(a^{2} + b^{2}\sin^{2}x) + c$

$b^{2}\log(a^{2} + b^{2}\sin^{2}x) + c$

$\int_{}^{}{\frac{1}{x\sqrt{1 + \log x}}\mspace{6mu} dx =}$.