Question

evaluate ₀∫²√x+1÷√x×dx

Answer 1

Answer:

Step-by-step explanation:

I

=

∫

(

x

+

1

)

√

2

−

x

d

x

Let  

u

=

2

−

x

. This implies that  

d

u

=

−

d

x

. Also note that  

−

u

=

x

−

2

, so  

−

u

+

3

=

x

+

1

. Then:

I

=

∫

(

−

u

+

3

)

√

u

(

−

d

u

)

=

∫

(

u

−

3

)

√

u

d

u

 

Expanding the square root as  

u

1

2

:

I

=

∫

(

u

(

u

1

2

)

−

3

u

1

2

)

d

u

=

∫

(

u

3

2

−

3

u

1

2

)

d

u

Now using  

∫

u

n

d

u

=

u

n

+

1

n

+

1

+

C

:

I

=

u

5

2

5

2

−

3

(

u

3

2

3

2

)

=

2

5

u

5

2

−

2

u

3

2

Factoring and making it look nice:

I

=

u

3

2

(

2

5

u

−

2

)

=

u

3

2

(

2

u

−

10

)

5

=

2

u

3

2

(

u

−

5

)

5

From  

u

=

2

−

x

:

I

=

2

(

2

−

x

)

3

2

(

(

2

−

x

)

−

5

)

5

=

−

2

(

2

−

x

)

3

2

(

x

+

3

)

5

+

C