Question

The integral $$\int {\frac{{dx}}{{{x^2}{{\left( {{x^4} + 1} \right)}^{\frac{3}{4}}}}}} $$   equals :

A. $$ - {\left( {{x^4} + 1} \right)^{\frac{1}{4}}} + c$$
B. $$ - {\left( {\frac{{{x^4} + 1}}{{{x^4}}}} \right)^{\frac{1}{4}}} + c$$  
C. $${\left( {\frac{{{x^4} + 1}}{{{x^4}}}} \right)^{\frac{1}{4}}} + c$$
D. $${\left( {{x^4} + 1} \right)^{\frac{1}{4}}} + c$$
Answer :   $$ - {\left( {\frac{{{x^4} + 1}}{{{x^4}}}} \right)^{\frac{1}{4}}} + c$$
Solution :
$$\eqalign{ & I = \int {\frac{{dx}}{{{x^2}{{\left( {{x^4} + 1} \right)}^{\frac{3}{4}}}}} = \int {\frac{{dx}}{{{x^3}{{\left( {1 + {x^{ - 4}}} \right)}^{\frac{3}{4}}}}}} } \cr & {\text{Let}}\,\,{x^{ - \,4}} = y \cr & \Rightarrow - 4{x^{ - \,3}}dx = dy\,\, \Rightarrow dx = \frac{{ - 1}}{4}{x^3}dy \cr & \therefore I = \frac{{ - 1}}{4}\int {\frac{{{x^3}dy}}{{{x^3}{{\left( {1 + y} \right)}^{\frac{3}{4}}}}}} \cr & = \frac{{ - 1}}{4}\int {\frac{{dy}}{{{{\left( {1 + y} \right)}^{\frac{3}{4}}}}}} \cr & = \frac{{ - 1}}{4} \times 4{\left( {1 + y} \right)^{\frac{1}{4}}} \cr & = - {\left( {1 + {x^{ - \,4}}} \right)^{\frac{1}{4}}} + C \cr & = - {\left( {\frac{{{x^4} + 1}}{{{x^4}}}} \right)^{\frac{1}{4}}} + C \cr} $$

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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