∫

5x+3

√x2+4x+10

dx
Now, 5x + 3 = A

d

dx

(x2+4x+10) + B
⇒ 5x + 3 = A (2x + 4) + B
⇒ 5x + 3 = 2Ax + 4A + B
⇒ 2A = 5 and 4A + B = 3
⇒ A =

5

2

Thus, 4 (

5

2

) + B = 3
⇒ 10 + B = 3
⇒ B = 3 - 10 = - 7
On substituting the values of A and B,we get
∫

5x+3

√x2+4x+10

dx = ∫

[ 5 2 d dx (x2+4x+10)−7]

√x2+4x+10

dx
= ∫ [

5 2 (2x+4)−7

√x2+4x+10

] dx
=

5

2

∫

2x+4

√x2+4x+10

dx - 7 ∫

dx

√x2+4x+10

=

5

2

I1−7I2 ... (1)
I1 = ∫

2x+4

√x2+4x+10

dx
Put x2 + 4x + 10 = z2
(2x + 4)dx = 2zdz
Thus, I1 = ∫

2z

z

dz = 2z = 2 √x2+4x+10+C1
I2 = ∫

dx

√x2+4x+10

= ∫

dx

√x2+4x+4+6

= ∫

dx

√(x+2)2+(√6)2

= log |(x + 2) + √x2+4x+10| + C2
Substituting I1 and I2 in(1),we get
∴ ∫

5x+3

√x2+4x+10

dx =

5

2

(2√x2+4x+10+C1) - 7 [log |(x + 2) + √x3+4x+10| + C2]
= 5 √x2+4x+10 - 7 [log |(x + 2) + √x3+4x+10|] +

5

2

C1−7C2
= 5 √x2+4x+10 - 7 [log |(x + 2) + √x3+4x+10|] + C , where C =

5

2

C1−7C2
OR
I = ∫

2x

√(x2+1)(x2+3)

Let x2 = z
∴ 2xdx = dz
∫ I = ∫

dz

(z+1)(z+3)

By partial fraction

1

(z+1)(z+3)

=

A

z+1

+

B

z+3

⇒ 1 = A (z + 3) + B (z + 1)
Putting z = - 3, we obtain :
1 = - 2B
B = -

1

2

∴ A =

1

2

∴

1

(z+1)(z+3)

=

1 2

z+1

+

(− 1 2 )

z+3

⇒ ∫

dz

(z+1)(z+3)

=

1

2

∫

dz

z+1

−

1

2

∫

dz

z+3

= 1/2 log |z + 1| -

1

2

log |z + 3| + C
∴ ∫

2xdx

(x2+1)(x2+3)

=

1

2

log |x2+1| -

1

2

log |x2+3| + C