if e 1 x 1 x dx x e 1 x 1 x 807028305

If $$\int {{{\log }_e}\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)} dx = x\,{\log _e}\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) + g\left( x \right) + C.$$
Then $$g\left( x \right) = ?$$

A. $$x - {\sin ^{ - 1}}x$$

B. $${\sin ^{ - 1}}x - x$$

C. $$x + {\sin ^{ - 1}}x$$

D. $${\sin ^{ - 1}}x - {x^2}$$

Answer : $${\sin ^{ - 1}}x - x$$

Solution :
$$I = \int {{{\log }_e}\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)} .1\,dx$$
Integrating by parts taking 1 as the second function.
$$\eqalign{ & I = \log \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)x - \int {\frac{1}{{\sqrt {1 - x} + \sqrt {1 + x} }}\left[ { - \frac{1}{{2\sqrt {1 - x} }} + \frac{1}{{2\sqrt {1 + x} }}} \right]\left( x \right)dx} \cr & \,\,\,\,\, = x\,\log \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) - \frac{1}{2}\int {\frac{{\sqrt {1 - x} - \sqrt {1 + x} }}{{\sqrt {1 - x} + \sqrt {1 + x} }}.\frac{1}{{\sqrt {1 - {x^2}} }}.x\,dx} \cr & \,\,\,\,\, = x\,\log \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) - \frac{1}{2}\int {\frac{{\left( {1 - x} \right) + \left( {1 + x} \right) - 2\sqrt {1 - {x^2}} }}{{\left( {1 - x} \right) - \left( {1 + x} \right)}}} .\frac{1}{{\sqrt {1 - {x^2}} }}.x\,dx \cr & \,\,\,\,\, = x\,\log \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) - \frac{1}{2}\int {\frac{{\sqrt {1 - {x^2}} - 1}}{{\sqrt {1 - {x^2}} }}} dx \cr & \,\,\,\,\, = x\,\log \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) - \frac{1}{2}\left[ {\int {1\,dx - \int {\frac{1}{{\sqrt {1 - {x^2}} }}} } dx} \right] \cr & \,\,\,\,\, = x\,\log \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) + \frac{1}{2}\left[ {{{\sin }^{ - 1}}x - x} \right] + C \cr & \therefore \,f\left( x \right) = x,\,g\left( x \right) = {\sin ^{ - 1}}x - x \cr} $$

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. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} + 5\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$

View Answer

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. $$\sqrt 3 $$

View Answer

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. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{2{x^2}}} + C$$

View Answer

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

View Answer

If $$\int {{{\log }_e}\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)} dx = x\,{\log _e}\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) + g\left( x \right) + C.$$
Then $$g\left( x \right) = ?$$

Releted MCQ Question on
Calculus >> Indefinite Integration

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

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

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

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-

Practice More Releted MCQ Question on
Indefinite Integration

Practice More MCQ Question on Maths Section

If $$\int {{{\log }_e}\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)} dx = x\,{\log _e}\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) + g\left( x \right) + C.$$ Then $$g\left( x \right) = ?$$

Releted MCQ Question on Calculus >> Indefinite Integration

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

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

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

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-

Practice More Releted MCQ Question on Indefinite Integration

Practice More MCQ Question on Maths Section

If $$\int {{{\log }_e}\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)} dx = x\,{\log _e}\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) + g\left( x \right) + C.$$
Then $$g\left( x \right) = ?$$

Releted MCQ Question on
Calculus >> Indefinite Integration

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-

Practice More Releted MCQ Question on
Indefinite Integration