Question

Integral of $f\left( x \right) = \sqrt {1 + {x^2}}$with respect to ${x^2}$ is
(A) $\dfrac{2}{3}\dfrac{{{{\left( {1 + {x^2}} \right)}^{\dfrac{3}{2}}}}}{x} + k$
(B) $\dfrac{2}{3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} + k$
(C) $\dfrac{2}{3}x{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} + k$
(D) none of these

Answer 1

Here we need to integrate the given function with respect to ${x^2}$ and the function is $f\left( x \right) = \sqrt {1 + {x^2}}$. So here we can integrate it directly by applying the formula or we can use substitution method and can substitute ${x^2}$ with t and then we can do further integration.
Formula used: $\int {{t^n}dt = \dfrac{{{t^{n + 1}}}}{{n + 1}} + k}$

Complete answer:
In the above question, we have
$f\left( x \right) = \sqrt {1 + {x^2}}$
So, we have to integrate the above function with respect to ${x^2}$
Let the required integral be equal to I.
$I\, = \,\int {\sqrt {1 + {x^2}} \,d{x^2}}$
Here, we had write $d{x^2}$ instead of dx because we have to integrate it with respect to ${x^2}$.
Let ${x^2} = t$
On substitution, we get
$I\, = \,\int {\sqrt {1 + t} \,dt}$
Now using the identity $\int {{t^n}dt = \dfrac{{{t^{n + 1}}}}{{n + 1}} + k}$
$I\, = \dfrac{{\,{{\left( {1 + t} \right)}^{\dfrac{1}{2} + 1}}}}{{\dfrac{1}{2} + 1}} + k$
$I = \dfrac{2}{3}{\left( {1 + t} \right)^{\dfrac{3}{2}}} + k$
Now, we will put the value of $t = {x^2}$ here
$I = \dfrac{2}{3}{\left( {1 + {x^2}} \right)^{\dfrac{3}{2}}} + k$

Therefore, the correct option is (B).

Note:
Here we should be careful while integrating the given function with respect to the variable because here we have to integrate it with respect to ${x^2}$ but not with respect to x. so, we should carefully use the formula here straight away and there is no need to do differentiation while substituting the given variable.