UPSC NDA I 2024 Mathematics Paper

To find the value of β, we need to solve the given integral and compare it with the provided solution form. The integral is:
∫‌

dx

√x+1−√x−1

Let's first simplify the integrand. To do this, multiply the numerator and the denominator by the conjugate of the denominator:
∫‌

√x+1+√x−1

(√x+1−√x−1)(√x+1+√x−1)

‌dx
The denominator simplifies to:
(√x+1)2−(√x−1)2=(x+1)−(x−1)=2
So the integral now becomes:
∫‌

√x+1+√x−1

2

‌dx=‌

1

2

‌∫(√x+1+√x−1)‌dx
This integral can be split into two separate integrals:
‌

1

2

(∫√x+1‌dx+∫√x−1‌dx)
For the first integral:
∫√x+1‌dx

Let u=x+1, hence du=dx. Then it becomes:
∫√udu=‌

2

3

u‌

3

2

=‌

2

3

(x+1)‌

3

2

For the second integral:
∫√x−1‌dx
Let v=x−1, hence dv=dx. Then it becomes:
∫√vdv=‌

2

3

v‌

3

2

=‌

2

3

(x−1)‌

3

2

Combining these results, we get:

‌

1

2

(‌

2

3

(x+1)‌

3

2

+‌

2

3

(x−1)‌

3

2

)=‌

1

3

(x+1)‌

3

2

+‌

1

3

(x−1)‌

3

2

+c
Comparing with the given form:
α(x+1)‌

3

2

+β(x−1)‌

3

2

+c
We see that:
α=‌

1

3

‌‌‌ and ‌‌‌β=‌

1

3