Question

If $$\int {\frac{{{x^2} - x + 1}}{{{x^2} + 1}}} {e^{{{\cot }^{ - 1}}x}}dx = A\left( x \right){e^{{{\cot }^{ - 1}}x}} + C,$$         then $$A\left( x \right)$$  is equal to :

A. $$ - x$$
B. $$x$$  
C. $$\sqrt {1 - x} $$
D. $$\sqrt {1 + x} $$
Answer :   $$x$$
Solution :
$$\eqalign{ & {\text{Let }}I = \int {\frac{{{x^2} - x + 1}}{{{x^2} + 1}}} .{e^{{{\cot }^{ - 1}}x}}dx \cr & {\text{Put }}x = \cot \,t \Rightarrow - {\text{cosec }}t\,dt = dx \cr & {\text{Now, }}1 + {\cot ^2} = {\text{cose}}{{\text{c}}^2}t \cr & \therefore \,I = \int {\frac{{{e^t}\left( {{{\cot }^2}t - \cot \,t + 1} \right)}}{{\left( {{\text{1}} + {{\cot }^2}t} \right)}}} \left( { - {\text{cose}}{{\text{c}}^2}t} \right)dt \cr & = \int {{e^t}\left( {\cot \,t - {\text{cose}}{{\text{c}}^2}t} \right)dt} \cr & = {e^t}\cot \,t + C \cr & = {e^{{{\cot }^{ - 1}}x}}\left( x \right) + C \equiv A\left( x \right).{e^{{{\cot }^{ - 1}}x}} + C \cr & \Rightarrow A\left( x \right) = x \cr} $$

Releted MCQ Question on
Calculus >> Indefinite Integration

Releted Question 1

The value of the integral $$\int {\frac{{{{\cos }^3}x + {{\cos }^5}x}}{{{{\sin }^2}x + {{\sin }^4}x}}dx} $$    is-

A. $$\sin \,x - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
B. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} + c$$
C. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
D. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} + 5\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
Releted Question 2

If $$\int_{\sin \,x}^1 {{t^2}f\left( t \right)dt = 1 - \sin \,x} ,$$      then $$f\left( {\frac{1}{{\sqrt 3 }}} \right)$$   is-

A. $$\frac{1}{3}$$
B. $${\frac{1}{{\sqrt 3 }}}$$
C. $$3$$
D. $$\sqrt 3 $$
Releted Question 3

Solve this $$\int {\frac{{{x^2} - 1}}{{{x^3}\sqrt {2{x^4} - 2{x^2} + 1} }}dx} = ?$$

A. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{{x^2}}} + C$$
B. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{{x^3}}} + C$$
C. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{x} + C$$
D. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{2{x^2}}} + C$$
Releted Question 4

Let $$I = \int {\frac{{{e^x}}}{{{e^{4x}} + {e^{2x}} + 1}}dx,\,J = \int {\frac{{{e^{ - \,x}}}}{{{e^{ - \,4x}} + {e^{ - \,2x}} + 1}}dx.} } $$
Then for an arbitrary constant $$C,$$ the value of $$J-I$$  equals-

A. $$\frac{1}{2}\log \left( {\frac{{{e^{4x}} - {e^{2x}} + 1}}{{{e^{4x}} + {e^{2x}} + 1}}} \right) + C$$
B. $$\frac{1}{2}\log \left( {\frac{{{e^{2x}} + {e^x} + 1}}{{{e^{2x}} - {e^x} + 1}}} \right) + C$$
C. $$\frac{1}{2}\log \left( {\frac{{{e^{2x}} - {e^x} + 1}}{{{e^{2x}} + {e^x} + 1}}} \right) + C$$
D. $$\frac{1}{2}\log \left( {\frac{{{e^{4x}} + {e^{2x}} + 1}}{{{e^{4x}} - {e^{2x}} + 1}}} \right) + C$$

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