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Math 109 Reviewer – A
...
2014-2015

∫ (x4 ▪ x-1/3 – x2 ▪ x-1/3 – 6 ▪ x-1/3 ) dx
∫ x11/3dx – ∫ x5/3dx – 6 ∫ x-1/3dx

Integration – process of a function whose derivative or
differential is given

11
5
1
n= 3
n=3
n =─ 3
3x14/3 3x8/3
2/3
14 – 8 – 9x + c

Integrand – the given function
Integral – the required function

THEOREM: Two functions having the same derivatives differ
at most by a constant

∫

3
...

CONSTANT INTEGRATION
PROPERTIES:

1
...

2
...


dy = (2x-5)dx

POWER FORMULAS:
xn+1
∫ x dx = n + 1 + c  if n ≠ –1
n

∫ x–1dx = lnx + c  if n = –1
EXAMPLES:
1
...


If dy = (2X – 5)dx and y = 2 when x = –1, find y when
x = 4
...
Y
...


c1 = -2
*substitute c1to the other equations to get the
other 2 constants

Find the equation of the curve if the slope at pt (2,3)
2x + 1
is given by 2y - 3
...


*substitute pt (2,3)

32 – 3(3) = 22 + 2 + c
c = -6
Equation: y2 – 3y = x2 + x – 6 (hyperbola)
3
...

*complete the square

x2 + 2x + __ = 8 – y + __
x2 + 2x + 1 = 9 – y
(x + 1)2 = –(y – 9)

d3y
If at any point (x,y) on a curve dx3 = 2 and (1,3) is the

*it is a parabola the opens
downward

V (-1,9)

pt
...

d3y
dx3 = 2
d
dx = 2
d = 2dx

dA = (yA – yB)dx
*dx = xLEFT – xRIGHT

dA = (8 – x2 – 2x)dx

∫dA = ∫ (8 – x2 – 2x)dx
x3
A = 8x – 3 – x2 + c

∫ d= ∫2dx
d2y
dx2 = 2x + c1

On the x-axis, y = 0
...

x3
*Substitute these values to A = 8x – 3 – x2 + c

dy 2
dx = x + c1x + c2

(-4)3
0 = 8(-4) – 3 – (-4)2 + c
80
c= 3

dy = (x2 + c1x + c2)dx

∫dy = ∫ (x2 + c1x + c2)dx
x3 c1x2
y = 3 + 2 + c2x + c3
SYSTEM OF EQUATIONS:
a
...
So we substitute it
to the last equation
...
slope = dx = -2 at x = 1
...

-2 = 12 + c1(1) + c2
c1 + c2 = -3
d2y
dx2 = 2x + c1
0 = 2 + c1`

NECES Academics Committee
Stephanie Grace de Guzman

80
When x = 2, and c = 3
x3
*Substitute these values to A = 8x – 3 – x2 + c
...


23
80
A = 8(2) – 3 – 22 + 3 = 36 sq
...


An art collector purchased for $1000 a painting by an
artist whose works are currently increasing with
respect to the time according to the formula
du
2/3
dt = 5t + 10t + 50

Math 109 Reviewer – A
...
2014-2015
*Substitute c1 and c2 to s = -16t2 + c1t + c2

where u dollars is the anticipated value of the
painting in t years after its purchase
...
Negative yung s
kasi opposite siya ng initial direction

-150 = -16t2 + 10t
16t2 – 10t – 150 = 0

+ 10t + 50)dt

*get t by using the quadratic formula

2

+ 5t + 50t + c

u = 3t5/3 + 5t2 + 50t + c
u
1000
0
t

t = 3
...
‘yung isa negative
...

*Substitute to u = 3t5/3 + 5t2 + 50t + c

ds
dt = -32t + 10

c = 1000
When t = 4 and c = 1000

When t = 3
...

v = -32(3
...
8 ft/s

*Substitute to u = 3t5/3 + 5t2 + 50t + c

u = $1,286
...


A woman in a hot air balloon dropped her binoculars
150ft above the ground and is rising at the rate of
10ft/s
...


b

∫
a

2
...


c

∫
a

b
c

f(x)dx + ∫ f(x)dx

b

∫
a

f(x)dx =

c

∫
a

b
c

f(t)dt + ∫ f(z)dz

EXAMPLES:
1
...
Subtract the lower
number from the upper number
...


5
1
1
28
3 (8 + 1) + 6 (4 – 1) - 2 (2+1) = 2 = 14
1 x3 + 1
∫ x + 1 dx
0

Math 109 Reviewer – A
...
2014-2015
(x + 1)(x2 - 2x + 1)
x+1

1

∫
0

1

∫
0

du = 8tdt
du
8 = tdt
1
n = -2

(x2 – 2x + 1)dx

1
x3
─ x2 + x
2
0
1 3
5
(1 – 0) – (12 – 0) + (1 – 0) = 6
3

]

* diba sa orig na formula it’s (4t2 + 9)-1/2tdt so diba u = (4t2 + 9)-1/2
tapos after that yung tdt
...
Diba nakuha nating du nung una is
8tdt
...
Pero gagawin siyang constant or “preparation” sa integration
...


∫ (x + 1)2dx

4
...


𝑡 𝑑𝑡

∫ (4t

2

3
-2
4 ∫ u du
3 u-1
4 ▪ -1 + c
3
─ 4(y4/3 + 9) + c
6
...
Pero yung 2 na trinanspose aka yung
1
1
2 , gagawin mong constant
...
So it
still follows the original formula na ∫und
...
=)))

3
...


∫
∫

∫ (1 + 2e3x)e3xdx
u = 1 + 2e3x
du = 6e3xdx

Math 109 Reviewer – A
...
2014-2015
du 3x
6 = e dx

2
─ u2 =

1
1
6 ∫ u du
1 u2
6▪ 2 +c
(1 + 2e3x)2
+c
12

7
...


∫

dx

𝑥4

4
1/2
3 ∫ u du
4 2 2/3
3▪3u +c
8 3/4
2/3
9 (x + 9) + c

𝑑𝑥
𝑥+√𝑥

𝑑𝑥

2√𝑥
𝑑𝑥

√𝑥

du
2 u = 2lnu + c
2ln( x + 1) + c

∫

3
...


2
1
( (10)−1)
4

1
...


1
4

EXAMPLES :

3

√ 4
𝑥 +9

(

2

2

─

10

]
𝑥−1)
2

4
...
To do that, substitute sa limits sa mga x sa
equation ng u which is 1 + 3ex
...
So 1 + 3(2) =
7
...
So 1 + 3(1) = 4
...
Y
...

6
...


1 du
3 u
1
x
3 ln(1 + 3e )
1
3 [ln7 – ln4]
1 7
3 ln4

∫

]

5
...


7

9
...


EXAMPLES:

∫ sin4xdx

1
...


26

∫ (x2 + 3x + 7 + x - 3 )dx
x3 3x2
-1
3 + 2 + 7x + 26∫ (x – 3) dx
x3 3x2
3 + 2 + 7x + 26ln(x – 3)
6
...


y+2
y2 + 4y dy

u = + 4y
du = (2y + 4)dy
du
2 = (y + 2)dy

2√𝑥
𝑑𝑥

2du =

u

1 -2 du 1
2 ∫ u = 2 ln |u|
-3
1
2 [ ln|-4| - ln|-3|]
1 4
2 ln3

√𝑥

2∫ tan u du = 2lnsec u + c
2ln(sec x ) + c

*change limits

-3
-3

dx

√𝑥

u= x
1
du =
dx

y2

x

𝑡𝑎𝑛√𝑥

-2
-4

]

∫ e2xcos e2x dx

3
...


∫ sinmx cosnx dx

∫ eudu = eu + c

where m or n is a positive odd integer
tools: change the one w/ odd powers
sin2x = 1 – cos2x
cos2x = 1 – sin2x

TRIGONOMETRIC FUNCTIONS
1
...

3
...


∫ cot u du = ln sin u + c
= – ln csc u + c

NECES Academics Committee
Stephanie Grace de Guzman

Ex:

∫ sin52x cos42x dx
y = 2x
dy = 2dx
dy
2 = dx

Math 109 Reviewer – A
...
2014-2015
1
5
4
2 ∫ sin y cos y dy
1
4
4
2 ∫ sin y cos y siny dy
1
2 2
4
2 ∫ (sin ) cos y siny dy
1
2 2
4
2 ∫ (1 – cos y) cos y siny dy
1
2
4
4
2 ∫ (1 – 2cos y + cos y) cos y siny dy
1
4
6
8
2 ∫ (cos y – 2cos y + cos y) siny dy
*integrate each term
...

u = cosy
du = -siny dy
un+1
So we’ll be using the form ∫un = n + 1 for each term
...


1
4
6
8
2 ∫ (u – 2u + u ) siny dy
1
–2 + c
1
–2 cos52x+ c
II
...
Where m is positive even integer
tools: sec2x = 1 + tan2x
csc2x = 1 + cot2x
Ex:
1
1
∫ tan42 x sec42 x dx
1
y=2x
2dy = dx

2∫ tan4y sec4y dy
2∫ tan4y sec2y sec2y dy
2∫ (tan4y + tan6y) sec2y dy
u = tany
du = sec2y
1

tan5 2 𝑥
𝟓

1

+

tan7 2 𝑥
7

)+c

b
...
∫ tannx dx or cotnx dx
where n is an integer
tools: tan2x = sec2x – 1
cot2x = csc2x – 1
a
...
Y
...
Use ∫undu
...


tan3x
– 3
tan3x
– 3
tan3x
– 3
tan3x
– 3

+ ∫ tan2x dx
+ ∫ (sec2x – 1) dx
+ ∫ sec2x dx – ∫dx
+ tan x – x

*combine na the two parts
...
n is a positive odd integer
EX:

∫ sin2x cos4x dx
∫ sin2x cos2x cos2x dx
∫ (sinx cosx)2 cos2x dx
1
∫ (2 sin2x)2cos2x dx

∫ tan5x dx
∫ tan3x ▪ tan2x dx
∫ tan3x(sec2x – 1) dx
∫ tan3x sec2x dx – ∫ tan3x dx
undu

*pwede na ma-integrate yung first term using
we’ll focus on the second term which is tan3xdx

tan4x
2
4 – ∫ tan x ▪ tanx dx
tan4x
2
4 – ∫ tanx(sec x – 1) dx
tan4x
2
4 – ∫ (tanx sec x – tanx) dx
tan4x
2
4 – ∫ (tanx sec x) dx – ∫ tanx dx
tan4x tan2x
4 – 2 – ln(secx) + c

IV
...
∫ sin ax sin bx dx

Math 109 Reviewer – A
...
2014-2015
1
2 ∫ (sinx cosx – sinx cos5x) dx

∫ sinmx cosnx dx
∫ sinmx cosnx dx
tools:

*for sinx cosx, u = sinx, du = cosxdx
...


1
sinα sinβ = 2 [cos(α ─ β) – cos(α + β)]
1
cosα cosβ = 2 [cos(α ─ β) + cos(α + β)]
1
sinα cosβ = 2 [sin(α ─ β) + sin(α + β)]

EX:
1
...


∫ sin4x sin7x dx
1
2 ∫ [cos(4x – 7x) – cos(4x + 7x)] dx
1
2 ∫ [cos(–3x) – cos(11x)] dx
1
2 ∫ (cos3x – cos11x) dx
1 1
1
2 [ 3 sin3x – 11 sin11x] + c
∫ cos7x sin4x dx

1 sin2x 1 1
2 ▪ 2 – 2 ∫ 2 [sin(x – 5x) + sin(x + 5x)] dx
sin2x 1
4 – 4 ∫ (-sin4x + sin6x) dx
sin2x 1
1
4 + 4 ∫ sin4x dx – 4 ∫ sin6x dx
sin2x
1
1
4 + 16 cos4x – 24 cos6x

1
2 ∫ [sin(4x – 7x) + sin(4x + 7x)] dx
1
2 ∫ [sin(–3x) + sin(11x)] dx
1
2 ∫ (–sin3x + sin11x) dx
1 1
1
2 [3 cos3x – 11 cos11x] + c

1
...

3
...


1
2 ∫ [cos(x – 3x) + cos(x + 3x)] dx
1
2 ∫ (cos2x + cos4x) dx

4
...


π
4

∫
∫
∫

a=5

u = 8x

du
8 = dx

a=3

u =2x

du
2 = dx

𝑑𝑥

√9−4𝑥 2
𝑑𝑥

1
2

du
a2 ─ u 2
1 -12x
2 Sin 3 + c

sinx sin2x sin3x dx

NECES Academics Committee
Stephanie Grace de Guzman

dx
25 + 64x2

√((3)2 −(2𝑥)2

]

1
2 ∫ sinx[cos(2x – 3x) – cos(2x + 3x)] dx
1
2 ∫ sinx[cosx – cos5x] dx

∫
∫
∫

1
du
8 a2 + u 2
1 1
8x
-1
8 ▪ 5 Tan 5 + c
1
8x
-1
40 Tan 5 + c

1 1
1
2 2 sin2x + 4 sin4x 0
1
2 –
1 1
1
1
2 2 (1) + 4 (0) – 0 = 4

]

∫
∫
∫

dx
(5)2 + (8x)2

cosx cos3x dx

[
[ ]
[

9
= 32

INVERSE TRIGONOMETRIC FUNCTIONS

π

∫4
0

0

Examples:

*let α = 4x and β = 7x

3
...


∫
∫

sec2 𝑥 𝑑𝑥
√50−sec2 𝑥
sec2 𝑥 𝑑𝑥
√50−(1+tan2 𝑥)

Math 109 Reviewer – A
...
2014-2015

∫
∫

1
sec2 𝑥 𝑑𝑥

sec2 𝑥 𝑑𝑥

a=7

√72 −(tan 𝑥)2
-1 𝒕𝒂𝒏 𝒙

u =tanx

du = sec2x dx

7
...


4
...


du = dx

Sin-1+ c

5
...


(x + 1)dx
x2 + 1

√5−2𝑥−3𝑥 2

22 1
formula c = 4(3) = 3

∫

u = 3x

x
*For x2 + 1 :

𝑑𝑥
b2
*to get c in ax2 + bx + c, get the value of 4a
...


𝑑𝑥
4
5

4
5

(5𝑥 2 −4𝑥+ )+(2− )
𝑑𝑥
(√5𝑥−

2 2
6 2
) +( )
√5
√5

du
a2 + u 2

1 √5

▪

Tan

√5 √6

a=

𝑑𝑢
= dx
√3

(

-1

5𝑥−2
√5
√6
√5

)+ c

NECES Academics Committee
Stephanie Grace de Guzman

2

√5

u=

5x─

2
√5

∫
1

∫
∫

dy
y(1 + ln2y)

dy
y(1 + (lny)2)
2

a=1

du
a2 + u 2

]

e

Tan-1(lny)
1
Tan-1(lne) – Tan-1(ln1)
Tan-11 – Tan-10
π
4

u = lny

dy
du = y

du
3 = dx

Math 109 Reviewer – A
...
2014-2015

4

10
...
So yung y sa labas, ihiwalay mo para
du
maging

...


∫
1

1

∫

𝑑𝑥
(𝑥+1)(√2)√𝑥(𝑥+2)

∫
∫
∫

√2

(2x - 4)dx
dx
x2 - 4x + 200 + 8 x2 - 4x + 200

1

2

u = x – 4x + 200
du = 2x – 4

√2

∫

1

∫

5
dx
2
2 ln(x – 4x + 20) + 8 (x - 2)2 + (4)2
a=4
u = x – 2 du = dx

2

u = 3 + 2x – x

√2
1
√2

]

6
2

𝑑𝑥
(𝑥+1)√(𝑥+1)2 −12

u=x+1

du = dx

a=1

]

2
1

[Sec-13 – Sec-12]
π
[Sec-13 – 3 ]

ADDITIONAL FORMULAS:
1
1
...


∫

3
...

du = (2 – 2x)dx

*divide the numerator by du
...
∫
1 √3+2𝑥−𝑥 2

𝑑𝑥
(𝑥+1)√2𝑥(𝑥+2)

∫

5
(2𝑥−4)+8
2
𝑥 2 −4𝑥+20

∫

2

2

2

*let u = x – 4x + 2z and
du = (2x – 4)dx
*divide the numerator by the derivative of the
denominator
...


u-1/2du + 4

3 2u1/2
─2 ▪ 1 + 4

∫

2

du = (2 – 2x)dx

∫

3
─2

du
2
u u2 ─ a 2
2 Sec-1 y

−2𝑥+2
∫√3+2𝑥−𝑥

3
─2

𝑑𝑦

5
...
Y
...


√(3𝑥 2 )2 −12

u = 3x2

du
6 = xdx

∫
∫

1−√2

4

5
...
So let’s represent

du = dx

a=

√3
2

Sin-1(

∫
∫(𝑥 −3𝑥+𝑑𝑥)−(10+ )
dx
x2 - 3x - 10
2

∫

3
u=x-2

3 2 49
(𝑥− ) −
2
4

4

du = dx

∫
0

3 7
2 2
3 7
(𝑥− )+
2 2

(𝑥− )−

7
a=2

ln
2√ 2

*note that we’re only getting (θ – β)
7

25

∫

π
4

cos 2𝑥
1
16

π
12

1
2

sin2 2𝑥−

u=x

du = dx

|]

𝑥+√2

1

𝑥−√2

0

NECES Academics Committee
Stephanie Grace de Guzman

a=

2

dx

1
a=4

u = sin2x

du
2 = cos2xdx

∫

du
u 2 - a2

4

|

5

(θ – β) = Sin-1( )

1 1
2 ▪ 2(1)ln

du
a2 - u 2

1

3

5

7
sin(θ – β) = 25

6
...


4

5

∫

1

5

sin(θ – β) = ( ) ( ) –( ) ( )

du
u 2 - a2

1
7 ln

3

)= θ and Sin-1( ) = β
...
Draw
4
ka ng triangle of each angle
...
So
the triangles would look like
this:

2

9
4

4

25
2 (θ – β)

1
ln|(x + 2 + x2 + x + 1 | + c

3
...


1−√2
1−√2

𝑥𝑑𝑥

2

0+√2

1+√2

√9𝑥 4 −1

1
6

1+√2

[ ln

ln

ln

|
|

|

1
4
1
sin 2𝑥+
4

sin 2𝑥−

𝜋 1
4
4
𝜋 1
sin 2( )+
4
4
𝜋 1
sin( )−
2
4
𝜋 1
sin( )+
2
4

|]

| |
| |

sin 2( )−

─ ln

─ ln

π
4
π
12

𝜋
1
12 4
𝜋
1
sin 2( )+
12 4
𝜋 1
sin( )−
6
4
𝜋 1
sin( )+
6
4

sin 2( )−

|

|

Math 109 Reviewer – A
...
2014-2015
3

1

ln ( 5 ) ─ ( 3 )
4

4
4

4

3
5

10
...


𝑥+1

1

∫

6y + 1
9y2 - 6y - 3 dy

∫
∫
∫

1
(18𝑦−6)+3
3
9𝑦 2 −6𝑦−3

u=

9y2

– 6y – 3

√𝑥

du = (18y – 6)dy

∫

1
3

du
du
u + 3 u 2 - a2

∫

du
u = 3y + 1 3 = dy

1
3
2
3 ln(9y – 6y – 3) + 4 ln

a=2

2
...


3y
| 3y +- 31 | + c

4
...

7
...

dx u = x2 + x + 2

du = (2x + 1)dx

*divide the numerator by du
...


∫
e^2
1
2

u=

1
1
√(𝑥 2 +𝑥+ )+(2− )
4
4

1-t

𝑑𝑡

─2du =

√1−𝑡

∫

─2 sech u ▪ tanh u ▪ du

√𝑢2 −𝑎2
1
– 4 ln|(x + 2 ) + x2 + x + 2 |} + c
1
(x + 2 ) +

lnx
x(ln4x - 1) dx

dx

√1−𝑡

𝑑𝑥

𝑑𝑢

2 x2 + x + 2 ─ 4 ln|

𝑑𝑥
2du =
a=1
√𝑥

sech(√1−𝑡) tanh(√1−𝑡)

dx

2

x

EXAMPLE:

∫

∫(2𝑥−1)−4
√𝑥 +𝑥+2
∫√𝑥2𝑥−1 ∫
+𝑥+2

u=

∫ sinh u du = cosh u + c
∫ cosh u du = sinh u + c
∫ tanh u du = ln |cosh u | +c
∫ coth u du = ln |sinh u | +c
∫ sech2 u du = tanh u + c
∫ csch2u du = –coth u + c
∫ sech u tanh u du = –sech u + c
∫ csch u coth u du = –csch u + c

1
...


∫√𝑥2𝑥−3
+𝑥+2

2

√(√ 𝑥) + 12 dx

HYPERBOLIC FUNCTIONS

(18y - 6)dy
dy
9y2 - 6y - 3 + 3 (9y2 - 6y + 1) - (3 + 1)

u

2 dx

2 ∫ u2 ± a2 du
1
2 { x ▪ x + 1 + ln | x + x + 1 |} + c

dy

1
3

∫

1
dx
𝑥

(√ 𝑥)

*divide the numerator by du
...


∫√1 +
∫√
∫

─2(–sech u) + c
2sech 1 - t + c

x2 + x + 2 |} + c

du lnx
u = ln2x 2 = x dx

a=1

IMPROPER INTEGRALS
I
...


∫

du
u 2 - a2

1 1
2 ▪ 2 ln
1
4

∞

ln
| ln xx +- 11 |]

|

2

2

e^3
𝑒^2

∫
a

b

f(x)dx = limb∞

b

∫
a

f(x)dx
b

∫ f(x)dx = lima -∞ ∫
-∞
a

f(x)dx



∞

| | 22 +- 11 |]

1
33 - 1
[ ln 33 + 1 ─ ln
4
1
4
3 1 4
4 [ ln5 ─ ln5 ] = 4 ln3

2

∫ f(x)dx = lima -∞ and b
-∞


2

NECES Academics Committee
Stephanie Grace de Guzman

NOTE:



b

∞ ∫ f(x)dx
a

Math 109 Reviewer – A
...
2014-2015
∞ 0
∞ & 0 = ‘pag ganyan yung situation, dun sa
equation/s kung sa’n naka substitute yung “b” or
“a”, derive both the numerator and the
denominator
...


b

limb∞

1
...
Integrals with infinite discontinuities in the integrand
*in other words, isa or both a and b sa formula
b
na ∫a f(x)dx, pag sinubstitute sa f(x)dx,
UNDEFINED yung lalabas
...
Babalik
tayo sa equation before this
...

Derive that
...


∫
0

∫

]

NECES Academics Committee
Stephanie Grace de Guzman

f(x)dx
b

f(x)dx + limmc+

∫
m

f(x)dx

𝑑𝑥
√𝑥(2−𝑥)

*pag sinubstite both 0 & 2, magiging undefined yung
sagot so ii-integrate both limits

𝑑𝑥

b

1 b u
─2
e du
1
b
1
─2
0
1
─2
1
─2
1
1
─2 ▪ ─1 = 2

b

EXAMPLES:

lima0 and b2
xe-x^2dx

∫
c

n

xe-x^2dx

∫
1

f(x)dx

c

f(x)dx =

= limnc-

1
1
─ 8 [–ln17] = 8 [ln17]

limb∞

∫
m

a) If f(x) increases numerically without limit as x  c,
a < c < b , (kumbaga yung point of discontinuity,
hindi given pero nasa gitna siya ng a and b) then,

∫
a

1
─ 8 [ln1 – ln17]

∞

f(x)dx

b

f(x)dx = limbn-

a
*recall that lnb = lna - lnb

∫
0

∫
a

a) If f(x) increases numerically without limit as x  b, then

1 1
─ 8 ln 17

2
...
Y
...
Evaluate:

8
...


∫ 3xex^2sin2ex^2 dx

1
...


∫
0

dx
(x - 1)2/3

2
...
But if you substitute 1, it will be undefined
...

1

∫
0

dx
(x - 1)2/3 +
b

limb1-

∫
0

2

∫
1

dx
(x - 1)2/3

(x – 2)-2/3dx + lima1+

]

3(x – 1)1/3

1/3

b
0

]

2

+ 3(x – 1)1/3

2

∫
a

3
...


(x – 2)-2/3dx
5
...

7
...
1
...
The curvature at any point (x,) on the curve: y’’= ______
b
...
The general equation of the curve: y = _______
If the curve has a critical point at (0,1) and the curve also
passes through (1,3):
d
...
At x = 1, y = _____
y’ = _____
2
...
Determine the
equations of the motion of the stone as functions of time
(Show the evaluation of the constants of integration):
a
...
velocity: v(t) = __________
c
...
time the stone takes before it hits the ground: t = ____s
e
...
Determine the area bounded by the curve y = x2 – 3x + 2
and the x-axis, from x = 0 to x = 2:
a
...
the intersections of the curve with the x-axis:
x1 = ____
x2 = ____
c
...
from x = x1 to x = 2 : c = ____
A = ____
e
...
Find the equation of the curve for which dx = x if the
curve passes through (1,2)
...


∫
∫
∫
∫
∫

17
...


11
...

13
...

15
...

19
...


∫
∫
∫

2

1 −
( 𝑥 3+

𝑥

3

)

√ 𝑥2 −1
−2
1
(𝑥 3 +√𝑥 2 −1)

dx

csc2(x2 – tanx) dx
sinh2(x2 - cosx)(2x + sinx)
dx
sech(x2 - cosx)
𝑥 3 +𝑥 2 −3√ 𝑥 2 +5
𝑥+1

dx

2x
(3x + 2)2 dx

e(2x + 1)
e(-5x - 2) dx
dx
x(1 + x2)
x+6
(x + 2)2 dx

Math 109 Reviewer – A
...
2014-2015

21
...


∫
1

7 - lnx
x(3 + lnx) dx

3

23
...

25
...

27
...

29
...


∫
∫

32
...


e2t
1 + 6e2t + 9e4t dx
cos(tanx3)
x-2 cos2x3 dx

34
...


∫ (sinx + tanx)2 dx

36
...


∫ [ sin 2

38
...


39
...

41
...


(1 + 6x2/3)
3 x + 6x5/3 dx
(4−tan 𝑥)
cos2 𝑥√4−tan2 𝑥

1
√(𝑥+1)(𝑥+10)

3y

∫
∫

dx

dx

y
y
+ cos2 ] cos2 dx
5 )(p3 + 5 )2
...
Y
...
1
...
36x2 + 6x + c1
b
...
3x4 + x3 + 2 + c2x + c3
d
...
y = 3
y’ = 11
2
...
-32
b
...
-16t2 + 16t
d
...
-80ft/s
x3 3x2
3
...
3 ─ 2 + 2x + c
b
...
c = 0

5
A=6

5
d
...
AT = 1 s
...

(lnx3)
4
...

1
...

3
...

5
...

7
...

9
...

11
...

13
...


2x2 + 4x + lnx + c
2x5/2 – 6x3/2 – 4x1/2 + c
2x5/2 2x3/2
5 + 3 +c
x3 + 3x2 + 15x + 25ln(x – 2) + c
1
2
-2x
2 ln(x + e ) + c
(x + x2)27
27 + c
ln(3 + lnx – x2) + c
1 (3π)-4x^2
─8 ln3π + c
3
─4 cos2ex^2 + c
1
2
x^2
2 ln(x + e ) + c
1 1/3
2
3
3 (x + x - 1 ) + c
tan(sinx + lnx) + x
─cot(x2 – tanx) + c
sinh3(x2 - cosx)
+c
3

NECES Academics Committee
Stephanie Grace de Guzman

x3
15
...
9 [ ln(3x + 2) + 3x + 2 ] + c
4 9x5 1
4
17
...
7 e7x + 3 + c
1
19
...
ln(x + 2) ─ x + 2
21
...
6
1
23
...
─ 6 ▪ cos23x + c
25
...
2 ln(1 + csc2y) + c
1
27
...


√4−cos 2𝑡
+
3

c

1
29
...
ln(1 + 3e2t) + c
1
31
...
2ln4 + c
33
...
4 Sin-1 4 ─ 4 - tan2x + c
1
1
35
...
3 Tan-1
+c
3

1
2

37
...


(𝑝3 +√5)

1
y 1
cos2y + 2 + 2 siny + c
4

4
...
9+3 ln 4

+c

1
2
2 ln(2 + tan x) + c
1 (Arccos2x)5
40
...

3 ─ 5 +c
1 3 5
1
sin32x
42
...




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