If ∫ ((x2 - x + 1/x2 + 1))⋅ e(cot)- 1 xdx =A(x)⋅ e(cot)- 1 x+C , where C is constant of integration, then A(x) is equal to

Question

If ∫ ((x2 - x + 1/x2 + 1))⋅ e(cot)- 1 xdx =A(x)⋅ e(cot)- 1 x+C , where C is constant of integration, then A(x) is equal to (A) -x (B) √1 - x (C) x (D) √1 + x

Tardigrade Admin · Accepted Answer

Method-I:
Differentiate both sides with respect to

x

, we get

\frac{( x ^{2} + 1 ) - x}{x ^{2} + 1} = \frac{( x ^{2} + 1 ) A ^{^{'}} ( x ) - A ( x )}{x ^{2} + 1}

∴ A (x) = x

Ans.
Method-II:
Let

I = \int (\frac{x ^{2} - x + 1}{1 + x ^{2}}) \cdot e^{(co t)^{- 1} x} d x

Put

co t^{- 1} x = t

, we get

I = \int e^{t} (co tt - (cosec)^{2} t) d t = e^{t} \cdot co t (t) + c

= x \cdot e^{co t^{- 1} x} + c

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

$- x$

$1 - x$

$x$

$1 + x$

Q. If ∫(x2+1x2−x+1​)⋅e(cot)−1xdx =A(x)⋅e(cot)−1x+C , where C is constant of integration, then A(x) is equal to

−x

1−x​

x

1+x​

Method-I: Differentiate both sides with respect to x , we get x2+1(x2+1)−x​=x2+1(x2+1)A′(x)−A(x)​ ∴A(x)=x Ans. Method-II: Let I=∫(1+x2x2−x+1​)⋅e(cot)−1xdx Put cot−1x=t , we get I=∫et(cott−(cosec)2t)dt=et⋅cot(t)+c =x⋅ecot−1x+c

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

$- x$

$1 - x$

$x$

$1 + x$