Question

Find the value of \[\int_{ - a}^a {\dfrac{1}{{x + {x^3}}}} dx\].
A \[0\]
B \[\int_0^a {\dfrac{1}{{1 + {x^6}}}dx} \]
C \[2\int_0^a {\dfrac{1}{{1 + {x^3}}}dx} \]
D \[\int_0^a {\dfrac{1}{{1 + {{\left( {a - x} \right)}^6}}}dx} \]

Answer 1

Hint: To evaluate the integral, check if function is odd or even. For all values of x, if the function \[f\]follows \[f\left( { - x} \right) = - f\left( x \right)\]then the function \[f\left( x \right)\]is odd. Then check where the function is continuous. If a real valued function is continuous at every point in its domain, then the function is said to be continuous. \[\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\] Then find the value of integral.





Complete step by step solution:The given integral is \[\int_{ - a}^a {\dfrac{1}{{x + {x^3}}}} dx\]. The function in the integral is \[\dfrac{1}{{x + {x^3}}}\].
First check the function is even or odd. Substitute \[x = - x\]
\[\begin{array}{l}f\left( x \right) = \dfrac{1}{{x + {x^3}}}\\f\left( { - x} \right) = \dfrac{1}{{\left( { - x} \right) + {{\left( { - x} \right)}^3}}}\end{array}\]
Simplify the function.
\[\begin{array}{l}f\left( { - x} \right) = \dfrac{1}{{ - x + ( - {x^3})}}\\f\left( { - x} \right) = - \left( {\dfrac{1}{{x + {x^3}}}} \right)\\f\left( { - x} \right) = - f\left( x \right)\end{array}\]
This shows that the function is odd.
Now check the continuity of the function at \[x = a\]. Substitute\[x = a\]in the function.
\[\begin{array}{l}f\left( x \right) = \dfrac{1}{{x + {x^3}}}\\f\left( a \right) = \dfrac{1}{{a + {a^3}}}\end{array}\]
Take limit of function at \[x = a\].
\[\mathop {\lim }\limits_{x \to a} f\left( x \right) = \mathop {\lim }\limits_{x \to a} \dfrac{1}{{x + {x^3}}}\]
Now evaluate the limit.
\[\begin{array}{l}\mathop {\lim }\limits_{x \to a} \dfrac{1}{{x + {x^3}}} = \dfrac{1}{{a + {a^3}}}\\\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\end{array}\]
So, we can say the function is continuous.
Integration of continuous odd function is always \[0\].
Hence, the value of \[\int_{ - a}^a {\dfrac{1}{{x + {x^3}}}} dx\] is zero.
\[\int_{ - a}^a {\dfrac{1}{{x + {x^3}}}} dx = 0\]



Option ‘A’ is correct



Note: The common mistake happen by students is considering the function is even and continuous that gives the value \[2\int_0^a {f\left( x \right)} \].