the integral 2 4 log x 2log x 2 log 36 12x x 2 dx is equal

The integral $$\int\limits_2^4 {\frac{{\log \,{x^2}}}{{\log \,{x^2} + \log \,\left( {36 - 12x + {x^2}} \right)}}} dx,$$ is equal to:

A. $$1$$

B. $$6$$

C. $$2$$

D. $$4$$

Answer : $$1$$

Solution :
$$\eqalign{ & I = \int\limits_2^4 {\frac{{\log \,{x^2}}}{{\log \,{x^2} + \log \,\left( {36 - 12x + {x^2}} \right)}}} \cr & I = \int\limits_2^4 {\frac{{\log \,{x^2}}}{{\log \,{x^2} + \log \,{{\left( {6 - x} \right)}^2}}}} .....(1) \cr & I = \int\limits_2^4 {\frac{{\log \,{{\left( {6 - x} \right)}^2}}}{{\log {{\left( {6 - x} \right)}^2} + \log \,{x^2}}}} .....(2) \cr & {\text{Adding (1) and (2), we get}} \cr & 2I = \int\limits_2^4 {dx} = \left[ x \right]_2^4 = 2\,\,\, \Rightarrow I = 1 \cr} $$

Releted MCQ Question on
Calculus >> Definite Integration

Releted Question 1

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-

A. $$ - 1$$

B. $$2$$

C. $$1 + {e^{ - 1}}$$

D. none of these

View Answer

Releted Question 2

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

A. no root in $$\left( {0,\,2} \right)$$

B. at least one root in $$\left( {0,\,2} \right)$$

C. a double root in $$\left( {0,\,2} \right)$$

D. two imaginary roots

View Answer

Releted Question 3

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-

A. $$\frac{\pi }{4}$$

B. $$\frac{\pi }{2}$$

C. $$\pi $$

D. none of these

View Answer

Releted Question 4

For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$ has the value-

A. $$\pi $$

B. $$1$$

C. $$0$$

D. none of these

View Answer

The integral $$\int\limits_2^4 {\frac{{\log \,{x^2}}}{{\log \,{x^2} + \log \,\left( {36 - 12x + {x^2}} \right)}}} dx,$$ is equal to:

Releted MCQ Question on
Calculus >> Definite Integration

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-

For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$ has the value-

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Practice More MCQ Question on Maths Section

The integral $$\int\limits_2^4 {\frac{{\log \,{x^2}}}{{\log \,{x^2} + \log \,\left( {36 - 12x + {x^2}} \right)}}} dx,$$ is equal to:

Releted MCQ Question on Calculus >> Definite Integration

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$ Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-

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Releted MCQ Question on
Calculus >> Definite Integration

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

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Definite Integration