Question

Let $f\left( x \right)$ be a polynomial of degree three such that $f\left( 0 \right) = 1$ , $f\left( 1 \right) = 2$ and $f\left( x \right)$ has a critical point at $x = 0$ where $f\left( x \right)$ does not have a local extremum, then $\int {\dfrac{{f\left( x \right)}}{{{x^2} + 1}}} dx$ is equal to
A. $x - \log \left( {{x^2} + 1} \right) + {\tan ^{ - 1}}x + c$
B. $x + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) - {\tan ^{ - 1}}x + c$
C. $\dfrac{1}{2}{x^2} + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) - {\tan ^{ - 1}}x + c$
D. $\dfrac{1}{2}\left( {{x^2} - \log \left( {{x^2} + 1} \right)} \right) + {\tan ^{ - 1}}x + c$

Answer 1

Hint: We can take $f\left( x \right)$ as $a{x^3} + b{x^2} + cx + d$ . Then by substituting the given values of the function we can find 2 equations. Then using the critical point, we will get an equation with the 1st derivative and with the 2nd derivative at the critical point, we get an equation as there is no local extremum. By solving the 4 equations we can find the values of a, b, c and d and find $f\left( x \right)$ . Now we can integrate the given function after simplifying and splitting the terms.

Complete step-by-step answer:
It is given that $f\left( x \right)$ be a polynomial of degree three. So, we can write the polynomial as,
 $f\left( x \right) = a{x^3} + b{x^2} + cx + d$ where a, b, c and d are constants.
We are given that $f\left( 0 \right) = 1$ . Applying this in the polynomial, we get,
 $f\left( 0 \right) = a \times {0^3} + b \times {0^2} + c \times 0 + d = 1$
 $ \Rightarrow d = 1$ … (1)
It is also given that $f\left( 1 \right) = 2$.
 $f\left( 1 \right) = a \times {1^3} + b \times {1^2} + c \times 1 + d = 2$
 $ \Rightarrow a + b + c + d = 2$ .. (2)
We can take the derivative of function $f\left( x \right)$ ,
 \[f'\left( x \right) = 3a{x^2} + 2bx + c\]
It is given that $f\left( x \right)$ has a critical point at $x = 0$
 $ \Rightarrow f'\left( 0 \right) = 0$
On substituting $x = 0$ on $f'\left( x \right)$ , we get,
 \[f'\left( 0 \right) = 3a \times {0^2} + 2b \times 0 + c = 0\]
On simplification, we get,
 $ \Rightarrow c = 0$ … (3)
We can take the second derivative of the function,
 \[ \Rightarrow f''\left( x \right) = 6ax + 2b\]
It is given that the function do not have a local extremum at $x = 0$ ,
 $ \Rightarrow f''\left( 0 \right) = 0$
Now we can substitute $x = 0$ in the 2nd derivative.
 \[ \Rightarrow f''\left( 0 \right) = 6a \times 0 + 2b = 0\]
On simplification, we get,
 $ \Rightarrow 2b = 0$
 $ \Rightarrow b = 0$ .. (4)
We can substitute equation (1), (3) and (4) in equation (2), we get,
 $ \Rightarrow a + 0 + 0 + 1 = 2$
On further calculations, we get,
 $ \Rightarrow a = 2 - 1 = 1$
Now we have the values of a, b, c and d. So the function $f\left( x \right)$ will become,
 $f\left( x \right) = {x^3} + 1$
Now we need to find the integral $\int {\dfrac{{f\left( x \right)}}{{{x^2} + 1}}} dx$ . On substituting the function, the integral will become
 $I = \int {\dfrac{{{x^3} + 1}}{{{x^2} + 1}}} dx$
We can add and subtract with x on the numerator.
 $ \Rightarrow I = \int {\dfrac{{{x^3} + x + 1 - x}}{{{x^2} + 1}}} dx$
Now we can take the common factor x from the 1st 2 terms of the numerator. So we get,
 $ \Rightarrow I = \int {\dfrac{{x\left( {{x^2} + 1} \right) + 1 - x}}{{{x^2} + 1}}} dx$
Now we can split the fraction
 $ \Rightarrow I = \int {\dfrac{{x\left( {{x^2} + 1} \right)}}{{{x^2} + 1}}} dx + \int {\dfrac{1}{{{x^2} + 1}}} dx - \int {\dfrac{x}{{{x^2} + 1}}} dx$
On simplification we get,
 $ \Rightarrow I = \int x dx + \int {\dfrac{1}{{{x^2} + 1}}} dx - \int {\dfrac{x}{{{x^2} + 1}}} dx$
We know that, $\int x dx = \dfrac{{{x^2}}}{2}$ and $\int {\dfrac{1}{{{x^2} + 1}}} dx = {\tan ^{ - 1}}x$ ….(A)
For the 3rd term, give substitution $u = {x^2} + 1$
Then by taking the derivative of u, we get,
 $\dfrac{{du}}{{dx}} = 2x$
On rearranging, we get,
 $ \Rightarrow xdx = \dfrac{{du}}{2}$
Giving the substitution, we get,
 $\int {\dfrac{x}{{{x^2} + 1}}} = \dfrac{1}{2}\int {\dfrac{{du}}{u}} $
We know that
 $ \Rightarrow \int {\dfrac{x}{{{x^2} + 1}}} = \dfrac{1}{2}\log u$
On substituting the value of u, we get,
 $ \Rightarrow \int {\dfrac{x}{{{x^2} + 1}}} = \dfrac{1}{2}\log \left( {{x^2} + 1} \right)$ ….(B)
Substituting values of (A) and (B) in $I = \int x dx + \int {\dfrac{1}{{{x^2} + 1}}} dx - \int {\dfrac{x}{{{x^2} + 1}}} dx$ ,
We get,
 $ \Rightarrow I = \dfrac{{{x^2}}}{2} + {\tan ^{ - 1}}x - \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + c$
We can rearrange it as $I = \dfrac{1}{2}\left( {{x^2} - \log \left( {{x^2} + 1} \right)} \right) + {\tan ^{ - 1}}x + c$
Therefore, the required integral is $\dfrac{1}{2}\left( {{x^2} - \log \left( {{x^2} + 1} \right)} \right) + {\tan ^{ - 1}}x + c$
So the correct answer is option D.

Note: Critical point of a function is the point at which the derivative of the function at that point is zero. Graphically, it is the point at which the graph of the function has a tangent which is parallel to the x axis. If the 2nd derivative at the critical point of function is greater than zero, it is the local minimum and if the 2nd derivative at the critical point of function is less than zero, it is the local maximum. If the 2nd derivative is equal to zero at the critical point, then it doesn't have a local maximum or minimum. For solving the integrating part, we must aware of the standard integrals such $\int {\dfrac{{dx}}{x}} = \log x$ , $\int x dx = \dfrac{{{x^2}}}{2}$ and $\int {\dfrac{1}{{{x^2} + 1}}} = {\tan ^{ - 1}}x$.