The integral $\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{\left(2-x^{2}\right) dx}{\left(2+x^{2}\right) \sqrt{4+x^{4}}}$ is equal to ____.

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Definite Integrals

The integral

\frac{24}{π} \int_{0}^{2} \frac{( 2 - x ^{2} ) d x}{( 2 + x ^{2} ) 4 + x ^{4}}

is equal to ____.

Answer: 3

Rewrite the integrand by dividing numerator and denominator by

x^{2}

I = \frac{24}{π} \int_{0}^{2} \frac{( \frac{2}{x ^{2}} - 1 ) d x}{( x + \frac{2}{x} ) ( x + \frac{2}{x} ) ^{2} - 4}

Substitute

t = x + \frac{2}{x}

, so

d t = (1 - \frac{2}{x ^{2}}) d x

, giving

- d t = (\frac{2}{x ^{2}} - 1) d x

.

The integral becomes:

I = - \frac{24}{π} \int \frac{d t}{t t ^{2} - 4} = - \frac{24}{π} \cdot \frac{1}{2} sec^{- 1} (\frac{t}{2}) + C

Evaluating from

x = 0

x = 2

: at

x = 2

t = 2 + 2 = 22

; as

x \to 0^{+}

t \to \infty

I = - \frac{12}{π} [sec^{- 1} (2) - sec^{- 1} (\infty)] = - \frac{12}{π} [\frac{π}{4} - \frac{π}{2}] = - \frac{12}{π} \cdot (- \frac{π}{4}) = 3

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Integer answer