1 x x 1 xe x 2dx is 488028351

$$\int {\frac{{\left( {1 + x} \right)}}{{x{{\left( {1 + x{e^x}} \right)}^2}}}dx} {\text{ is }} = ?$$

A. $$\ln \left| {\frac{{x{e^x}}}{{1 + x{e^x}}}} \right| + \frac{1}{{1 + x{e^x}}} + C$$

B. $$\left( {1 + x{e^x}} \right) + \ln \left| {\frac{{x{e^x}}}{{1 + x{e^x}}}} \right| + C$$

C. $$\frac{1}{{1 + x{e^x}}} + \ln \left| {x{e^x}\left( {1 + x{e^x}} \right)} \right| + C$$

D. none of these

Answer : $$\ln \left| {\frac{{x{e^x}}}{{1 + x{e^x}}}} \right| + \frac{1}{{1 + x{e^x}}} + C$$

Solution :
$$\eqalign{ & I = \int {\frac{{\left( {1 + x} \right)}}{{x{{\left( {1 + x{e^x}} \right)}^2}}}dx} \cr & \,\,\,\,\, = \int {\frac{{\left( {1 + x} \right){e^x}}}{{x{e^x}{{\left( {1 + x{e^x}} \right)}^2}}}dx} \cr & {\text{Put }}x{e^x} = t \Rightarrow \left( {{e^x} + x{e^x}} \right)dx = dt \cr & I = \int {\frac{{dt}}{{t{{\left( {1 + t} \right)}^2}}}} \cr & {\text{Let }}\frac{1}{{t{{\left( {1 + t} \right)}^2}}} = \frac{A}{t} + \frac{B}{{1 + t}} + \frac{D}{{{\left( {1 + t} \right)}^2}},{\text{ we get}} \cr & A = \frac{1}{{{{\left( {1 + 0} \right)}^2}}} = 1,\,\,D = \frac{1}{{ - 1}} = - 1 \cr & {\text{Equating coefficient of }}{t^2},\,\,\,0 = A + B \Rightarrow B = - 1 \cr & \therefore \,I = \int {\left[ {\frac{1}{t} - \frac{1}{{1 + t}} - \frac{1}{{{{\left( {1 + t} \right)}^2}}}} \right]} dt \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \ln \left| t \right| - \ln \left| {1 + t} \right| + \frac{1}{{1 + t}} + C \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \ln \left| {\frac{t}{{1 + t}}} \right| + \frac{1}{{1 + t}} + C \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \ln \left| {\frac{{x{e^x}}}{{1 + x{e^x}}}} \right| + \frac{1}{{1 + x{e^x}}} + C \cr} $$

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

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. $$\sqrt 3 $$

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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

$$\int {\frac{{\left( {1 + x} \right)}}{{x{{\left( {1 + x{e^x}} \right)}^2}}}dx} {\text{ is }} = ?$$

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

$$\int {\frac{{\left( {1 + x} \right)}}{{x{{\left( {1 + x{e^x}} \right)}^2}}}dx} {\text{ is }} = ?$$

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

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