6.

But by example (3)

7.f

=

22 √ x2± a2 F

4

(x + √ x2 ± a2) + C

= x23 √ x2 + a2 + 3a2,

3a2

dx

VEITS VE()

u =

X3

1. Let

x

u' = z vr±a2 7/log. (x + √x2 ± a2)

√ x2± a2±

26 dx

√ x2 + a2 = √x2 + a2

2. Let

When m is even

du =

When m is odd

Similarly

ax

4. a2. x2

4 2 a2

.

"SV==@= √== "" { ++++++

a2

a2

5.3

5 3 1

·

·

log. (x + √x2 + a2) + C x dx

8. x2±a2

x2 + a2

® √ √ = ± @ = V Z ± e { Z = 6.4.2

7.5

And generally, if

x x

√ x2 + a2

...

(m-1) (m 3). m (m-1) (m — 4)

Next, to integrate

du =

m

-

u = √ x2± a2

3.1 xam 4.2

1) (m m. (m

du =
Va2.

α

u = sin.-1

...

dx

X

α

xdx √a2-x2

xm-1

m

3.1.a2

4.2.1

{

3.1.2.x
4.2

4.2.am-1
5.3.1

S

xm-1

m

3)
2)

5. a2. 2-3

6.4

+

+

...

+

3.1.a1
4.2

} + c

+

3u

2)

5.3.1.a
4.
2

5.3.a1.x 6.4.

6.4.a. x2
7.5.3

x dx

√ a2 — x2

then by VI. Chap. I.

3.1.

+
m (m

log.(x+√x2+a3)+C

(m—1)x1—3a2 (m—1) (m—3)x1—5a*—

m (m

· 2) (m —

x2

+C, the radius being unity.

then u = - √a2x2 + C.

a

4.2

C

1.a

5.3. ± 2 S 6.4.2

log.

6.4.2.a +7.5.

(m—1)xm—3α2, (m—1) (m—3)xm-5α.

±

m (m-2) ·m (m-2) (m

·4) +

log. (x+a+a2) + C

+

4)

+

3. Let

Let

..

du =

1. Let

Let

=

u =

2. Let

To integrate

du=

y=

==

du =

Integrating by parts,

y =

du =

dy

a2 +

2=

=

du =

√ a2 + x2

a2 x

Integrating by parts

y2

dx

xm Va2 x2

The process is precisely analogous to that employed in the last case

dx

a2 x

- y3 dy Vy2-I

dx

1
x+1

dy

་

ydy √y — 1

3. 1 ર 3.1

{-2}- Ad log. (1+VI+) 4. 2.x2 4.2

Let

2.

In like manner all integrals of the form

dx

xVx2 -a

t

du = x2 dx √ a2 + x2

=

4x+

x dx Va2x2, xTM dx √x2 - a2.

which are
both known forms.

1

3

= — (a2 + z2) } —

4

3.y

+ 3 : 2 } +

.

dx

x

% (@2 + x2) ; — — √ √ Œ2 + ** = { - 4 S

4

which are both known forms.

log.(y+√y®—1)+C

3.1

ર

++ 1}+ 3-log. (1 + √ 122)+Cc 4.2.2*3+

3.1

4.2

a1

4.2

Multiply both numerator and denominator by √a2 + x2

x2 dx (a2 + x2)

Va2 + x2

= log. (x + V@++} + ©

2

1

{

3

The preceding integral may be found in a manner somewhat different

du = x2 dx va2 + x2

a2

x2 dx

+ f Ja2 + a2

4

log. (x +√a2+x2) + C

3. Let

Let

..

du =

1. Let

Let

du=

2. Let

To integrate

y=

du =

Integrating by parts,

du =

dx

x5 √1 + x2

ydy

u = f − y3 • √ i + y2

Integrating by parts

Va2 + x2

a2 x

dy

du =

y=

X

du == dy

dx

xm Va2

The process is precisely analogous to that employed in the last case

dx XVI-2

== √1 + y2 { 2/2 - 3:3} - 12 log. (y +√l + 3o)

3. у
4.2

3.1
4.2

4

22

dx

dx

x2

- y3dy =

dy= and — yo dy =

dx

Let

1.

may be determined.

To integrate

- y dy

In like manner all integrals of the form

dx

xmx2

integrating by parts

=

du = x2 dx √ a2 + x2

x

— (a2 + x^

3

x dx Va2x2,

=

x

= % («2 + x2)

4

which are both known forms.

Ꮖ

= — (0° + 29

(a2 3

3

3

which are both known forms.

2,

du x3 dx va2 = x22

4.2.

1

3

— ƒ dx (a2 + x2) √

at

4

a2 x

4.2.

a2

Multiply both numerator and denominator by a2 + x2

x2 dx (a2 + x2)

√ a2 + x2

+ x2

น

log. (x + √ a2 + x2)} + c

The preceding integral may be found in a manner somewhat different

du = x2 dx √ a2 + x2

a2

a1

.log. (x + √ a* +2°) +C

4.2