First slide

Methods of integration

Question

Let $f (x) = \frac{1}{x^{6} + x^{4}}$ and $F$ be a antiderivative
of $f$ such that $F (1) = \frac{π}{4} + \frac{2}{3}$
Statement-1 $F (\frac{1}{\sqrt{3}}) = \frac{π}{6}$
Statement-2: $F (x) = \tan^{- 1} x + \frac{1}{3 x} - \frac{1}{x^{3}}$

Moderate

A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for
STATEMENT-1
C

STATEMENT-1 is True, STATEMENT-2 is False
D

STATEMENT-1 is False, STATEMENT-2 is True

Solution

$f (x) = \frac{1}{x^{4} (x^{2} + 1)} = \frac{1}{x^{4}} - \frac{1}{x^{2}} + \frac{1}{x^{2} + 1}$
So $\begin{array}{l} F (x) = - \frac{1}{3 x^{3}} + \frac{1}{x} + \tan^{- 1} x + C \end{array}$
$\begin{aligned} F (1) = \frac{2}{3} + \frac{π}{4} + C \end{aligned}$ so $\begin{aligned} C = 0 \end{aligned}$
$\begin{aligned} F (x) = \tan^{- 1} x + \frac{1}{x} - \frac{1}{3 x^{3}} \Rightarrow F (\frac{1}{\sqrt{3}}) = \frac{π}{6} . \end{aligned}$

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