First slide

Methods of integration

Question

Let $f (x)$ be a polynomial of degree three
such that $f (0) = 1, f (1) = 2 and 0$ is a critical point of $f (x)$
such that $f (x)$ does not have a local extremum at 0. Then
$\int \frac{f (x)}{x^{2} + 1} d x$ is equal to

Moderate

A

$x - \log (x^{2} + 1) + \tan^{- 1} x + C$
B

$x + (1 / 2) \log (x^{2} + 1) - \tan^{- 1} x + C$
C

$(1 / 2) x^{2} + (1 / 2) \log (x^{2} + 1) - \tan^{- 1} x + C$
D

$(1 / 2) (x^{2} - \log (x^{2} + 1)) + \tan^{- 1} x + C$

Solution

Let $f (x) = a x^{3} + b x^{2} + c x + d .$ Since
$f (0) = 1$ so $d = 1 .$ Moreover, $f (1) = 2$
$a + b + c = 1$ since 0 is a critical point, 0 $= f^{'} (0) = 3 a .0$
$+ 2 b .0 + C$ .i.e., $C = 0 .$ Also $f^{''} (x) = 6 a x + 2 b$ and since
$f (x)$ does not have extremum at 0 so 0 $= f^{''} (0) = b$ .
Hence $a = 1$ and
$\begin{array}{l} \int \frac{f (x)}{x^{2} + 1} d x = \int \frac{x^{3} + 1}{x^{2} + 1} d x = \int (x - \frac{1}{2} \frac{2 x}{x^{2} + 1} + \frac{1}{x^{2} + 1}) d x \\ = \frac{1}{2} [x^{2} - \log (x^{2} + 1)] + \tan^{- 1} x + C . \end{array}$

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