If I=∫ (x2-x+1/x2+1).ecot-1xdx= A(x)ecot-1x+C, then A(x) is equal to :

Question

If I=∫ (x2-x+1/x2+1).ecot-1xdx= A(x)ecot-1x+C, then A(x) is equal to : (A) -x (B) x (C) √1-x (D) √1+x

Tardigrade Admin · Accepted Answer

Let I=∫ (x2-x+1/x2+1).ecot-1xdx Put x=cot t ⇒ -cosec2 dt=dx Now, 1 + cot2 t = cosec2 t ∴ I=∫ (et(cot2 t-cot t+1)/(1+cot2 t))(-cosec2 t)dt =- ∫ et(cosec2 t-cot t)dt =∫ et(cot t-coesc2 t)dt =et cot t+C =ecot-1x(x)+C≡ A(x).ecot-1x+C

Q. If $I = \int \frac{x ^{2} - x + 1}{x ^{2} + 1} . e^{co t^{- 1} x} d x = A (x) e^{co t^{- 1} x} + C,$ then $A (x)$ is equal to :

$- x$

$x$

$1 - x$

$1 + x$

Q. If I=∫x2+1x2−x+1​.ecot−1xdx=A(x)ecot−1x+C, then A(x) is equal to :

−x

x

1−x​

1+x​

Let I=∫x2+1x2−x+1​.ecot−1xdx Put x=cott⇒−cosec2dt=dx Now, 1+cot2t=cosec2t ∴I=∫(1+cot2t)et(cot2t−cott+1)​(−cosec2t)dt =−∫et(cosec2t−cott)dt =∫et(cott−coesc2t)dt =etcott+C =ecot−1x(x)+C≡A(x).ecot−1x+C

Q. If $I = \int \frac{x ^{2} - x + 1}{x ^{2} + 1} . e^{co t^{- 1} x} d x = A (x) e^{co t^{- 1} x} + C,$ then $A (x)$ is equal to :

$- x$

$x$

$1 - x$

$1 + x$