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JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Definite Integrals

If

\int_{0}^{3} \frac{15 x ^{3}}{1 + x ^{2} + ( 1 + x ^{2} ) ^{3}} dx = α 2 + β 3

, where

α, β

are integers, then

α + β

is equal to ____.

Answer: 10

Solution

Substitute

1 + x^{2} = t^{2}

, so

2 x d x = 2 t d t

, giving

x d x = t d t

.

When

x = 0

,

t = 1

; when

x = 3

,

t = 2

.

The integral becomes:

\int_{1}^{2} \frac{15 ( t ^{2} - 1 ) t d t}{t t ^{2} + t ^{3}} = 15 \int_{1}^{2} \frac{t ^{2} - 1}{t 1 + t} d t

Substitute

1 + t = u^{2}

, so

d t = 2 u d u

:

15 \int_{2}^{3} \frac{( u ^{2} - 1 ) ^{2} - 1}{u} \cdot 2 u d u = 30 \int_{2}^{3} (u^{4} - 2 u^{2}) d u

= 30 [\frac{u ^{5}}{5} - \frac{2 u ^{3}}{3}]_{2}^{3}

= 30 [\frac{9 3 - 4 2}{5} - \frac{2 ( 3 3 - 2 2 )}{3}]

= 30 [\frac{- 3}{5} + \frac{8 2}{15}]

= - 63 + 162

Therefore

α = 16

,

β = - 6

, and

α + β = 10

.

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Integer answer

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