Question

Solve this : $$\int {{x^{51}}\left( {{{\tan }^{ - 1}}x + {{\cot }^{ - 1}}x} \right)dx} = ?$$

A. $$\frac{{{x^{52}}}}{{52}}\left( {{{\tan }^{ - 1}}x + {{\cot }^{ - 1}}x} \right) + c$$  
B. $$\frac{{{x^{52}}}}{{52}}\left( {{{\tan }^{ - 1}}x - {{\cot }^{ - 1}}x} \right) + c$$
C. $$\frac{{\pi {x^{52}}}}{{104}} + \frac{\pi }{2} + c$$
D. $$\frac{{{x^{52}}}}{{52}} + \frac{\pi }{2} + c$$
Answer :   $$\frac{{{x^{52}}}}{{52}}\left( {{{\tan }^{ - 1}}x + {{\cot }^{ - 1}}x} \right) + c$$
Solution :
$$\eqalign{ & \int {{x^{51}}\left( {{{\tan }^{ - 1}}x + {{\cot }^{ - 1}}x} \right)dx} \cr & = \int {{x^{51}}.\frac{\pi }{2}dx} \,\,\,\,\,\left\{ {\therefore \,{{\tan }^{ - 1}}x + {{\cot }^{ - 1}}x = \frac{\pi }{2}} \right\} \cr & = \frac{{\pi {x^{52}}}}{{104}} + c \cr & = \frac{{{x^{52}}}}{{52}}\left( {{{\tan }^{ - 1}}x + {{\cot }^{ - 1}}x} \right) + c \cr} $$

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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Indefinite Integration


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