Question

$\frac{3x - 4}{3x + 4}$ = t 3x – 4 = 3xt + 4t x = $\frac{4t + 4}{3(1 - t)}$ f(t) = $\frac{4t + 4}{3(1 - t)}$ + 2 f(x) = $\frac{4x + 4}{3(1 - x)}$ + 2 = $\frac{4(x - 1) + 8}{3(1 - x)}$ + 2

f(x) = 2 – $\frac{4}{3}$ – $\frac{8}{3(x - 1)}$ = $\frac { 2 } { 3 } - \frac { 8 } { 3 ( x - 1 ) }$ $\int_{}^{}{f(x)dx}$ = $\frac{2}{3}$x – $\frac{8}{3}$ln |x – 1| + c

• A. $\frac{\cos 2x}{2}$+ c
• B. sin 2x + c
• C. – sinx + c
• D. None

Answer 1

Answer: (C)

– sinx + c

Answer 2

0 < x < p/2 , 0 < sinx < 1

$\lim_{n \rightarrow \infty}$ (sin x)ⁿ ® 0

f(x) = $\frac{(\sin x)^{2n} - 1}{(\sin x)^{2n} + 1}$ = – 1

= – sin x + c