AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Consider the expression,
I=∫

x

√x+1+√x−1

dx
=∫

x[√x+1−√x−1]

x+1−x+1

dx
=

1

2

‌∫x√x+1dx−

1

2

‌∫x√x−1dx
=

1

2

I1−

1

2

I2
Then,
I1=∫x√x+1dx
Put x+1=u⇒dx=du
I1=∫(u−1)√udx
=∫(u

3

2

−u

1

2

)dx
=

2

5

u

5

2

−

2

3

u

3

2

+c1
=

2

5

(x+1)

5

2

−

2

3

(x+1)

3

2

+c1
Further simplify the above,
I1=2(x+1)

3

2

[

3x+3−5

15

]+c1
=

2(3x−2)

15

(x+1)

3

2

+c1
Now,
I2=∫x√x−1dx
Put x−1=v⇒dx=dv
I2=∫(v+1)√vdv
=∫(v

3

2

+v

1

2

)dv
=

2

5

v

5

2

+

2

3

v

3

2

+c2
=2(x−1)

3

2

[

1

5

(x−1)+

1

3

]+c2
=2(x−1)

3

2

[

3x−3+5

15

]+c2
Further simplify the above,
I2=

2(3x+2)

15

(x−1)

3

2

+c2
This gives,
I=

1

2

I1−

1

2

I2
=

(3x−2)

15

(x+1)

3

2

−

(3x+2)

15

(x−1)

3

2

+c [∵c=c1+c2]
This implies,
A(x)=

1

15

(3x−2) and B(x)=

−1

15

(3x+2)
So,
A(x)+B(x)=

1

15

(3x−2)−

1

15

(3x+2)
=

−4

15