Question

$a^{x}\log a + c$

• A. $\log a + c$
• B. $a^{x} + c$
• C. $\int_{}^{}{\sec x\tan xdx =}$
• D. $\sec x + \tan x + c$

Answer 1

Answer: (A)

$\log a + c$

Answer 2

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

$\frac{1}{b}(a + b\cos x) + c$

{Multiplying $\frac{1}{b(a + b\cos x)} + c$ and $\frac{1}{b}\log(a + b\cos x) + c$ by $\int_{}^{}{\frac{1}{x^{3}}\lbrack\log x^{x}\rbrack^{2}\mspace{6mu} dx =}$

Now putting $\frac{x^{3}}{3}(\log x) + x + c$

we get the integral

$\frac{1}{3}(\log x)^{3} + c$

Trick : Since $3\log(\log x) + c$

$\int_{}^{}{\frac{\mathbf{1}}{\mathbf{x}}\mathbf{\sec}^{\mathbf{2}}\mathbf{(}\mathbf{\log}\mathbf{x}\mathbf{)dx =}}$