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Question: How do you find the antiderivative of \(\dfrac{{5{x^2}}}{{{x^2} + 1}}\)?...

How do you find the antiderivative of 5x2x2+1\dfrac{{5{x^2}}}{{{x^2} + 1}}?

Explanation

Solution

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′=fF' = 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 1x2\dfrac{1}{{{x^2}}}. Assume 1x2=t2\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 1x2=t2\dfrac{1}{{{x^2}}} = {t^2}.

Complete step-by-step solution:
In this question, we want to find antiderivative of
⇒∫5x2x2+1dx\Rightarrow \int {\dfrac{{5{x^2}}}{{{x^2} + 1}}dx}
Let us divide the denominator by x2{x^2}.
⇒∫5x2x2(1+1x2)dx\Rightarrow \int {\dfrac{{5{x^2}}}{{{x^2}\left( {1 + \dfrac{1}{{{x^2}}}} \right)}}dx}
Cut x2{x^2} as it is the common factor in the numerator and in the denominator.
⇒∫51+1x2dx\Rightarrow \int {\dfrac{5}{{1 + \dfrac{1}{{{x^2}}}}}dx} ……………....(1)
Now, let us assume1x2=t2\dfrac{1}{{{x^2}}} = {t^2}
Applying square roots on both sides,
x=1tx = \dfrac{1}{t}
That is equal to
⇒x=t−1\Rightarrow x = {t^{ - 1}}
Let us find a derivative of the above expression.
⇒dxdt=d(t−1)dt\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{{d\left( {{t^{ - 1}}} \right)}}{{dt}}
Now, apply the formula
ddxxn=n×xn−1\dfrac{d}{{dx}}{x^n} = n \times {x^{n - 1}}
That is equal to
⇒dxdt=−1×t−1−1\Rightarrow \dfrac{{dx}}{{dt}} = - 1 \times {t^{ - 1 - 1}}
So,
⇒dxdt=−t−2\Rightarrow \dfrac{{dx}}{{dt}} = - {t^{ - 2}}
That is equal to
⇒dxdt=−1t2\Rightarrow \dfrac{{dx}}{{dt}} = - \dfrac{1}{{{t^2}}}
Therefore,
dx=(−1t2)dtdx = \left( { - \dfrac{1}{{{t^2}}}} \right)dt
Let us put the value of dx in the equation (1).
⇒∫51+t2×(−1t2)dt\Rightarrow \int {\dfrac{5}{{1 + {t^2}}} \times \left( { - \dfrac{1}{{{t^2}}}} \right)dt}
Take constant terms out-side the integration.
⇒−5×∫1t2(1+t2)dt\Rightarrow - 5 \times \int {\dfrac{1}{{{t^2}\left( {1 + {t^2}} \right)}}dt}
In the numerator, we can replace 1 by 1+t2−t21 + {t^2} - {t^2}.
⇒−5[∫1+t2−t2t2(1+t2)]dt\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.
⇒−5[∫1+t2t2(1+t2)−t2t2(1+t2)]dt\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,
⇒−5[∫1t2−11+t2]dt\Rightarrow - 5\left[ {\int {\dfrac{1}{{{t^2}}} - \dfrac{1}{{1 + {t^2}}}} } \right]dt
Apply integration individually.
⇒−5[∫1t2dt−∫11+t2dt]\Rightarrow - 5\left[ {\int {\dfrac{1}{{{t^2}}}dt - \int {\dfrac{1}{{1 + {t^2}}}} dt} } \right]
Let us apply the formula ∫t−2dt=t−1−1\int {{t^{ - 2}}dt} = \dfrac{{{t^{ - 1}}}}{{ - 1}} and ⇒∫11+t2dt=arctan⁡(t) \Rightarrow \int {\dfrac{1}{{1 + {t^2}}}} dt = \arctan \left( t \right)
⇒−5[t−1−1−arctan⁡(t)]\Rightarrow - 5\left[ {\dfrac{{{t^{ - 1}}}}{{ - 1}} - \arctan \left( t \right)} \right]
⇒5t−5arctan⁡(t)+c\Rightarrow \dfrac{5}{t} - 5\arctan \left( t \right) + c
Let us take x=1tx = \dfrac{1}{t}
That is equal to t=1xt = \dfrac{1}{x}
⇒5x−5arctan⁡(1x)+c\Rightarrow 5x - 5\arctan \left( {\dfrac{1}{x}} \right) + c

The required answer for the given question is 5x−5arctan⁡(1x)+c5x - 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