Question

If $\int \frac{xcos\alpha +1}{(x^2+2xcos\alpha +1{)}^{3/2}}$ $dx=\frac{x}{\sqrt{f\left(x\right)+g\left(x\right)cos\alpha }}+c$ then (more than one option is correct)

• A. f(1) = 2
• B. f(2) = 5
• C. g(1) = 2
• D. g(2) = 2

Answer 1

Answer: (A), (B), (C)

The correct options are
A f(1) = 2
B f(2) = 5
C g(1) = 2

put $[1+\frac{2cos\alpha }{x}+\frac{1}{x^2}]=t^2$

We get

$\int \frac{xcos\alpha +1}{(x^2+2xcos\alpha +1{)}^{3/2}}$ $dx=\frac{x}{\sqrt{x^2+1+2xcos\alpha }}+c$

ie $f\left(x\right)=x^2+1$

$g\left(x\right)=2x$