Question

How do you find the antiderivative of $\dfrac{{5{x^2}}}{{{x^2} + 1}}$?

Answer 1

Hint: An anti-derivative of a function f is a function whose derivative is f. In other words, F is an anti-derivative of f if $F' = f$, to find an antiderivative for a function f, we can often reverse the process of differentiation. First, convert the expression in the form of $\dfrac{1}{{{x^2}}}$. Assume $\dfrac{1}{{{x^2}}} = {t^2}$ . find the differentiation in terms of y. Substitute the values. And find integration. Then again convert the answer in terms of x by putting $\dfrac{1}{{{x^2}}} = {t^2}$.

Complete step-by-step solution:
In this question, we want to find antiderivative of
$ \Rightarrow \int {\dfrac{{5{x^2}}}{{{x^2} + 1}}dx} $
Let us divide the denominator by ${x^2}$.
$ \Rightarrow \int {\dfrac{{5{x^2}}}{{{x^2}\left( {1 + \dfrac{1}{{{x^2}}}} \right)}}dx} $
Cut ${x^2}$ as it is the common factor in the numerator and in the denominator.
 $ \Rightarrow \int {\dfrac{5}{{1 + \dfrac{1}{{{x^2}}}}}dx} $ ……………....(1)
Now, let us assume$\dfrac{1}{{{x^2}}} = {t^2}$
Applying square roots on both sides,
$x = \dfrac{1}{t}$
That is equal to
$ \Rightarrow x = {t^{ - 1}}$
Let us find a derivative of the above expression.
$ \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{{d\left( {{t^{ - 1}}} \right)}}{{dt}}$
Now, apply the formula
$\dfrac{d}{{dx}}{x^n} = n \times {x^{n - 1}}$
That is equal to
$ \Rightarrow \dfrac{{dx}}{{dt}} = - 1 \times {t^{ - 1 - 1}}$
So,
$ \Rightarrow \dfrac{{dx}}{{dt}} = - {t^{ - 2}}$
That is equal to
$ \Rightarrow \dfrac{{dx}}{{dt}} = - \dfrac{1}{{{t^2}}}$
Therefore,
$dx = \left( { - \dfrac{1}{{{t^2}}}} \right)dt$
Let us put the value of dx in the equation (1).
$ \Rightarrow \int {\dfrac{5}{{1 + {t^2}}} \times \left( { - \dfrac{1}{{{t^2}}}} \right)dt} $
Take constant terms out-side the integration.
$ \Rightarrow - 5 \times \int {\dfrac{1}{{{t^2}\left( {1 + {t^2}} \right)}}dt} $
In the numerator, we can replace 1 by $1 + {t^2} - {t^2}$.
$ \Rightarrow - 5\left[ {\int {\dfrac{{1 + {t^2} - {t^2}}}{{{t^2}\left( {1 + {t^2}} \right)}}} } \right]dt$
Now, let us split the expression.
$ \Rightarrow - 5\left[ {\int {\dfrac{{1 + {t^2}}}{{{t^2}\left( {1 + {t^2}} \right)}} - \dfrac{{{t^2}}}{{{t^2}\left( {1 + {t^2}} \right)}}} } \right]dt$
That is equal to,
$ \Rightarrow - 5\left[ {\int {\dfrac{1}{{{t^2}}} - \dfrac{1}{{1 + {t^2}}}} } \right]dt$
Apply integration individually.
$ \Rightarrow - 5\left[ {\int {\dfrac{1}{{{t^2}}}dt - \int {\dfrac{1}{{1 + {t^2}}}} dt} } \right]$
Let us apply the formula $\int {{t^{ - 2}}dt} = \dfrac{{{t^{ - 1}}}}{{ - 1}}$ and $ \Rightarrow \int {\dfrac{1}{{1 + {t^2}}}} dt = \arctan \left( t \right)$
$ \Rightarrow - 5\left[ {\dfrac{{{t^{ - 1}}}}{{ - 1}} - \arctan \left( t \right)} \right]$
$ \Rightarrow \dfrac{5}{t} - 5\arctan \left( t \right) + c$
Let us take $x = \dfrac{1}{t}$
That is equal to $t = \dfrac{1}{x}$
$ \Rightarrow 5x - 5\arctan \left( {\dfrac{1}{x}} \right) + c$

The required answer for the given question is $5x - 5\arctan \left( {\dfrac{1}{x}} \right) + c$.

Note: To find antiderivatives of elementary functions is harder than finding their derivatives because there is no predefined method for computing indefinite integrals.
There are many properties and techniques for finding antiderivatives:
The linearity of integration that breaks complicated integrals into simpler ones.
 Integration by substitution
The inverse chain rule method
Integration by parts
Inverse function integration
The method of partial fraction integration.
Numerical integration