Indefinite Integral
See how to solve an indefinite integral.
25 examples and their solutions.
Integral: Definition
Definition
Integral
Derivative
Integral and derivative are the opposite operations.Derivative
f'(x) is the derivative of f(x).
So f(x) is the integral of f'(x).
This is why integral is also called antiderivative.
Integral
Derivative
So, finding ∫ f(x) dx is finding the function whose derivative is f(x).Derivative
∫ f(x) dx is read as [integral f x d x].
Integral of a⋅f(x) + b⋅g(x)
Property
∫[af(x) + bg(x)] dx
= a∫f(x) dx + b∫g(x) dx
= a∫f(x) dx + b∫g(x) dx
Integral of xn
Formula
∫xn dx = 1n + 1xn + 1 + C
xnWrite the reciprocal of (n + 1): 1/(n + 1).
Write xn + 1.
Don't forget to write the constant term +C.
The derivative of [1/(n + 1)]xn + 1 + C is xn.
→ Integral of xn is [1/(n + 1)]xn + 1 + C.
Derivative of xn
Example
∫x3 dx
Solution ∫x3 dx
= 14x4 + C - [1]
= 14x4 + C - [1]
[1]
x3
3 + 1 = 4
Write the reciprocal of 4: 1/4.
Write x4.
Write +C.
3 + 1 = 4
Write the reciprocal of 4: 1/4.
Write x4.
Write +C.
Close
Example
∫ dx
Solution ∫ dx
= ∫1 dx
= ∫x0 dx - [1]
= 1x1 + C - [2]
= x + C - [3]
= ∫1 dx
= ∫x0 dx - [1]
= 1x1 + C - [2]
= x + C - [3]
[2]
x0
0 + 1 = 1
Write the reciprocal of 1: 1.
Write x1.
Write +C.
0 + 1 = 1
Write the reciprocal of 1: 1.
Write x1.
Write +C.
[3]
∫ dx = x + C
∫ a dx = ax + C
∫ a dx = ax + C
Close
Example
∫(6x2 - 2x + 5) dx
Solution ∫(6x2 - 2x + 5) dx
= 6⋅13x3 - 2⋅12x2 + 5x + C - [1]
= 2x3 - x2 + 5x + C
= 6⋅13x3 - 2⋅12x2 + 5x + C - [1]
= 2x3 - x2 + 5x + C
[1]
x2
2 + 1 = 3
Write the reciprocal of 3: 1/3.
Write x3.
x = x1
1 + 1 = 2
Write the reciprocal of 2: 1/2.
Write x2.
+5 = 5⋅x0
0 + 1 = 1
Write +5.
Write the reciprocal of 1: 1.
Write x1.
2 + 1 = 3
Write the reciprocal of 3: 1/3.
Write x3.
x = x1
1 + 1 = 2
Write the reciprocal of 2: 1/2.
Write x2.
+5 = 5⋅x0
0 + 1 = 1
Write +5.
Write the reciprocal of 1: 1.
Write x1.
Close
Example
∫√x dx
Solution ∫√x dx
= ∫x12 dx
= 23x32 + C - [1]
= 23x1 + 12 + C
= 23x√x + C
= ∫x12 dx
= 23x32 + C - [1]
= 23x1 + 12 + C
= 23x√x + C
[1]
x1/2
1/2 + 1 = 3/2
Write the reciprocal of 3/2: 2/3.
Write x3/2.
Write +C.
1/2 + 1 = 3/2
Write the reciprocal of 3/2: 2/3.
Write x3/2.
Write +C.
Close
Integral of sin x
Formula
∫sin x dx = -cos x + C
Derivative of cos x Example
∫(sin x2 + cos x2)2 dx
Solution ∫(sin x2 + cos x2)2 dx
= ∫(sin2 x2 + 2 sin x2 cos x2 + cos2 x2) dx - [1]
= ∫(1 + sin x) dx - [2]
= x - cos x + C
= ∫(sin2 x2 + 2 sin x2 cos x2 + cos2 x2) dx - [1]
= ∫(1 + sin x) dx - [2]
= x - cos x + C
[1]
[2]
Close
Integral of cos x
Formula
∫cos x dx = sin x + C
Derivative of sin x Example
∫cos2 x2 dx
Solution ∫cos2 x2 dx
= ∫1 + cos x2 dx - [1]
= 12∫(1 + cos x) dx
= 12(x + sin x) + C
= ∫1 + cos x2 dx - [1]
= 12∫(1 + cos x) dx
= 12(x + sin x) + C
Close
Integral of sec2 x
Formula
∫sec2 x dx = tan x + C
Derivative of tan x Example
∫cos 2xcos2 x dx
Solution ∫cos 2xcos2 x dx
= ∫2cos2 x - 1cos2 x dx - [1]
= ∫(2 - 1cos2 x) dx
= ∫(2 - sec2 x) dx - [2]
= 2x - tan x + C
= ∫2cos2 x - 1cos2 x dx - [1]
= ∫(2 - 1cos2 x) dx
= ∫(2 - sec2 x) dx - [2]
= 2x - tan x + C
[2]
1/cos x = sec x
Trigonometry (Right Triangle)
Trigonometry (Right Triangle)
Close
Integral of ex
Formula
∫ex dx = ex + C
Derivative of ex Example
∫e2x - 1ex - 1 dx
Solution ∫e2x - 1ex - 1 dx
= ∫(ex + 1)(ex - 1)ex - 1 dx
= ∫(ex + 1) dx
= ex + x + C
= ∫(ex + 1)(ex - 1)ex - 1 dx
= ∫(ex + 1) dx
= ex + x + C
Close
Integral of ax
Formula
∫ax dx = 1ln aax + C
Example
∫2x(3x - 1) dx
Solution ∫2x(3x - 1) dx
= ∫(6x - 2x) dx - [1]
= 1ln 66x - 1ln 22x + C
= ∫(6x - 2x) dx - [1]
= 1ln 66x - 1ln 22x + C
[1]
Close
Integral of 1x
Formula
∫1x dx = ln |x| + C
Example
∫x2 - x + 1x dx
Solution ∫x2 - x + 1x dx
= ∫(x - 1 + 1x) dx
= 12x2 - x + ln |x| + C
= ∫(x - 1 + 1x) dx
= 12x2 - x + ln |x| + C
Close
Integral by Substitution
Example
∫(2x - 1)8 dx
Solution ∫(2x - 1)8 dx
2x - 1 = t - [1]
2 dx = dt - [2]
dx = 12dt - [3]
= ∫t8⋅12dt - [4]
= 12∫t8 dt
= 12⋅19t9 + C
= 118(2x - 1)9 + C - [5]
2x - 1 = t - [1]
2 dx = dt - [2]
dx = 12dt - [3]
= ∫t8⋅12dt - [4]
= 12∫t8 dt
= 12⋅19t9 + C
= 118(2x - 1)9 + C - [5]
[1]
If 2x - 1 = t,
you can solve ∫ t8.
→ Set 2x - 1 = t.
you can solve ∫ t8.
→ Set 2x - 1 = t.
[2]
Differentiate both sides.
Write dx after differentiating the x term.
and write dt after differentiating the t term.
Write dx after differentiating the x term.
and write dt after differentiating the t term.
[3]
Change the equation to [dx = ...].
[4]
2x - 1 → t
dx → [1/2]dt
dx → [1/2]dt
[5]
t → 2x - 1
Close
Example
∫sin5 x cos x dx
Solution ∫sin5 x cos x dx
sin x = t
cos x dx = dt - [1] [2]
= ∫t5 dt - [3]
= 16t6 + C
= 16sin6 x + C - [4]
sin x = t
cos x dx = dt - [1] [2]
= ∫t5 dt - [3]
= 16t6 + C
= 16sin6 x + C - [4]
[2]
The given expression has cos x dx.
So let's use this.
So let's use this.
[3]
sin x → t
cos x dx → dt
cos x dx → dt
[4]
t → sin x
Close
Integral of tan x
Formula
∫tan x dx = -ln |cos x| + C
Example
∫tan x dx
Solution ∫tan x dx
= ∫sin xcos x dx
cos x = t
-sin x dx = dt - [1]
sin x dx = -dt - [2]
= ∫1t (-dt) - [3]
= -∫1t dt
= -ln |t| + C - [4]
= -ln |cos x| + C - [5]
= ∫sin xcos x dx
cos x = t
-sin x dx = dt - [1]
sin x dx = -dt - [2]
= ∫1t (-dt) - [3]
= -∫1t dt
= -ln |t| + C - [4]
= -ln |cos x| + C - [5]
[2]
The given expression has sin x dx.
So let's use this.
So let's use this.
[3]
[5]
t → cos x
Close
Integral of f(ax + b)
Formula
∫f(ax + b) dx = 1aF(ax + b) + C
F(x): Integral of f(x) 1/a: Reciprocal of a
Example
∫(2x - 1)8 dx
Solution ∫(2x - 1)8 dx
= 12⋅19(2x - 1)9 + C
= 118(2x - 1)9 + C
= 12⋅19(2x - 1)9 + C
= 118(2x - 1)9 + C
Close
Example
Integral of a Rational Expression
Example
∫x2 - 3x + 5x - 1 dx
Solution ∫x2 - 3x + 5x - 1 dx
1-35
11-2
1-23
= ∫(x - 2 + 3x - 1) dx - [1]
= 12x2 - 2x + 3 ln |x - 1| + C - [2] [3]
1-35
11-2
1-23
= ∫(x - 2 + 3x - 1) dx - [1]
= 12x2 - 2x + 3 ln |x - 1| + C - [2] [3]
[3]
Close
Example
∫3x - 2x(x - 1) dx
Solution ∫3x - 2x(x - 1) dx
3x - 2x(x - 1) = Ax + Bx - 1 - [1]
= A(x - 1)x(x - 1) + Bxx(x - 1)
= A(x - 1) + Bxx(x - 1)
= Ax - A + Bxx(x - 1)
= (A + B)x - Ax(x - 1)
A + B = 3
-A = -2 - [2]
A = 2
2 + B = 3
B = 1
= ∫(2x + 1x - 1) dx - [3]
= 2 ln |x| + ln |x - 1| + C
3x - 2x(x - 1) = Ax + Bx - 1 - [1]
= A(x - 1)x(x - 1) + Bxx(x - 1)
= A(x - 1) + Bxx(x - 1)
= Ax - A + Bxx(x - 1)
= (A + B)x - Ax(x - 1)
A + B = 3
-A = -2 - [2]
A = 2
2 + B = 3
B = 1
= ∫(2x + 1x - 1) dx - [3]
= 2 ln |x| + ln |x - 1| + C
[1]
Set (3x - 2)/(x(x - 1)) = A/x + B/(x - 1).
Partial Fraction Decompoistion
Partial Fraction Decompoistion
[2]
(A + B)x = 3x
-A = -2
-A = -2
[3]
A = 2, B = 1
(3x - 2)/(x(x - 1)) = A/x + B/(x - 1)
= 2/x + 1/(x - 1)
(3x - 2)/(x(x - 1)) = A/x + B/(x - 1)
= 2/x + 1/(x - 1)
Close
Trigonometric Substitution
Formula
√a2 - x2 → x = a sin θ
a2 + x2 → x = a tan θ
Special case of integral by substitution a2 + x2 → x = a tan θ
Example
∫dx√a2 - x2
Solution ∫dx√a2 - x2
x = a sin θ
dx = a cos θ dθ - [1]
a sin θ = x
sin θ = xa
θ = arcsin xa - [2]
= ∫a cos θ dθ√a2 - a2 sin2 θ
= ∫a cos θ dθ√a2(1 - sin2 θ)
= ∫a cos θ dθ√a2cos2 θ - [3]
= ∫a cos θ dθa cos θ - [4]
= ∫dθ
= θ + C - [5]
= arcsin xa + C - [6]
x = a sin θ
dx = a cos θ dθ - [1]
a sin θ = x
sin θ = xa
θ = arcsin xa - [2]
= ∫a cos θ dθ√a2 - a2 sin2 θ
= ∫a cos θ dθ√a2(1 - sin2 θ)
= ∫a cos θ dθ√a2cos2 θ - [3]
= ∫a cos θ dθa cos θ - [4]
= ∫dθ
= θ + C - [5]
= arcsin xa + C - [6]
[2]
[5]
∫ dθ = ∫ 1 dθ
= θ + C
= θ + C
[6]
θ → arcsin x/a
Close
Example
∫dxa2 + x2
Solution ∫dxa2 + x2
x = a tan θ
dx = a sec2 θ dθ - [1]
a tan θ = x
tan θ = xa
θ = arctan xa - [2]
= ∫a sec2 θ dθa2 + a2 tan2 θ
= ∫a sec2 θ dθa2(1 + tan2 θ)
= ∫sec2 θ dθa sec2 θ - [3]
= ∫dθa
= 1a∫ dθ
= 1aθ + C
= 1aarctan xa + C
x = a tan θ
dx = a sec2 θ dθ - [1]
a tan θ = x
tan θ = xa
θ = arctan xa - [2]
= ∫a sec2 θ dθa2 + a2 tan2 θ
= ∫a sec2 θ dθa2(1 + tan2 θ)
= ∫sec2 θ dθa sec2 θ - [3]
= ∫dθa
= 1a∫ dθ
= 1aθ + C
= 1aarctan xa + C
[2]
[3]
1 + tan2 θ = sec2 θ
Trigonometry Formula
Trigonometry Formula
Close
Integral by Parts
Formula
∫uv' dx = uv - ∫u'v dx
Proof [uv]' = u'v + uv' - [1]
uv' + u'v = [uv]'
uv' = [uv]' - u'v
∫uv' = uv - ∫u'v - [2]
uv' + u'v = [uv]'
uv' = [uv]' - u'v
∫uv' = uv - ∫u'v - [2]
[2]
Integral both sides.
How to Determine u and v'
ln x, loga x
x, x2, x3, ...
sin x, cos x
ex, ax u↕v'
Upper function → u x, x2, x3, ...
sin x, cos x
ex, ax u↕v'
(= The integral is complex.)
Lower function → v'
(= The integral is simple.)
Example
∫xex dx
Solution u v'
∫xex dx
u = x v = ex
u' = 1 v' = ex - [1]
= xex - ∫1⋅ex dx - [2]
= xex - ∫ex dx
= xex - ex + C
∫xex dx
u = x v = ex
u' = 1 v' = ex - [1]
= xex - ∫1⋅ex dx - [2]
= xex - ∫ex dx
= xex - ex + C
[1]
Write u = x.
→ u' = x
Derivative Rules
Write v' = ex next to u' = 1.
→ Write v = ex next to u = x.
Integral of ex
→ u' = x
Derivative Rules
Write v' = ex next to u' = 1.
→ Write v = ex next to u = x.
Integral of ex
[2]
uv: xex
u'v: 1⋅ex
u'v: 1⋅ex
Close
Example
∫x cos x dx
Solution u v'
∫x cos x dx
u = x v = sin x
u' = 1 v' = cos x - [1]
= x sin x - ∫1⋅sin x dx - [2]
= x sin x - ∫sin x dx
= x sin x - (-cos x) + C - [3]
= x sin x + cos x + C
∫x cos x dx
u = x v = sin x
u' = 1 v' = cos x - [1]
= x sin x - ∫1⋅sin x dx - [2]
= x sin x - ∫sin x dx
= x sin x - (-cos x) + C - [3]
= x sin x + cos x + C
[1]
Write u = x.
→ u' = x
Write v' = cos x next to u' = 1.
→ Write v = sin x next to u = x.
Integral of cos x
→ u' = x
Write v' = cos x next to u' = 1.
→ Write v = sin x next to u = x.
Integral of cos x
[2]
uv: x sin x
u'v: 1⋅sin x
u'v: 1⋅sin x
Close
Example
∫x2 sin x dx
Solution u v'
∫x2 sin x dx
u = x2 v = -cos x
u' = 2x v' = sin x - [1]
= x2 (-cos x) - ∫2x⋅(-cos x) dx - [2]
= -x2 cos x + 2∫x cos x dx - [3]
u = x v = sin x
u' = 1 v' = cos x
= -x2 cos x + 2⋅[x sin x - ∫1⋅sin x dx]
= -x2 cos x + 2[x sin x - ∫sin x dx]
= -x2 cos x + 2[x sin x - (-cos x) + C]
= -x2 cos x + 2[x sin x + cos x + C]
= -x2 cos x + 2x sin x + 2 cos x + C
= 2x sin x + (-x2 + 2) cos x + C
∫x2 sin x dx
u = x2 v = -cos x
u' = 2x v' = sin x - [1]
= x2 (-cos x) - ∫2x⋅(-cos x) dx - [2]
= -x2 cos x + 2∫x cos x dx - [3]
u = x v = sin x
u' = 1 v' = cos x
= -x2 cos x + 2⋅[x sin x - ∫1⋅sin x dx]
= -x2 cos x + 2[x sin x - ∫sin x dx]
= -x2 cos x + 2[x sin x - (-cos x) + C]
= -x2 cos x + 2[x sin x + cos x + C]
= -x2 cos x + 2x sin x + 2 cos x + C
= 2x sin x + (-x2 + 2) cos x + C
[1]
Write u = x2.
→ u' = 2x
Write v' = sin x next to u' = 2x.
→ Write v = -cos x next to u = x2.
→ u' = 2x
Write v' = sin x next to u' = 2x.
→ Write v = -cos x next to u = x2.
[2]
uv: x2 (-cos x)
u'v: 2x⋅(-cos x)
u'v: 2x⋅(-cos x)
[3]
Do the integral by parts again.
Close
Example
∫ex sin x dx
Solution v' u
∫ex sin x dx
u = sin x v = ex
u' = cos x v' = ex
= (sin x)⋅ex - ∫(cos x)⋅ex dx - [1]
= ex sin x - ∫(cos x)⋅ex dx - [2]
u = cos x v = ex
u' = -sin x v' = ex
= ex sin x - [(cos x)⋅ex - ∫(-sin x)⋅ex dx] - [3]
= ex sin x - [ex cos x + ∫ex sin x dx]
= ex sin x - ex cos x - ∫ex sin x dx
(given) = ex sin x - ex cos x - (given) - [4]
2(given) = ex sin x - ex cos x + C
= ex(sin x - cos x) + C
∴ (given) = ex2(sin x - cos x) + C
∫ex sin x dx
u = sin x v = ex
u' = cos x v' = ex
= (sin x)⋅ex - ∫(cos x)⋅ex dx - [1]
= ex sin x - ∫(cos x)⋅ex dx - [2]
u = cos x v = ex
u' = -sin x v' = ex
= ex sin x - [(cos x)⋅ex - ∫(-sin x)⋅ex dx] - [3]
= ex sin x - [ex cos x + ∫ex sin x dx]
= ex sin x - ex cos x - ∫ex sin x dx
(given) = ex sin x - ex cos x - (given) - [4]
2(given) = ex sin x - ex cos x + C
= ex(sin x - cos x) + C
∴ (given) = ex2(sin x - cos x) + C
[1]
uv: (sin x)⋅ex
u'v: (cos x)⋅ex
u'v: (cos x)⋅ex
[2]
Do the integral by parts again.
[3]
uv: (cos x)⋅ex
u'v: (-sin x)⋅ex
u'v: (-sin x)⋅ex
[4]
∫ ex sin x dx = (given)
Close
Integral of ln x
Formula
∫ln x dx = x ln x - x + C
Example
∫ln x dx
Solution ∫ln x dx
u v'
= ∫(ln x)⋅1 dx - [1]
u = ln x v = x
u' = 1x v' = 1 - [2]
= (ln x)⋅x - ∫ 1x⋅x dx - [3]
= x ln x - ∫1 dx
= x ln x - x + C
u v'
= ∫(ln x)⋅1 dx - [1]
u = ln x v = x
u' = 1x v' = 1 - [2]
= (ln x)⋅x - ∫ 1x⋅x dx - [3]
= x ln x - ∫1 dx
= x ln x - x + C
[2]
Write u = ln x.
→ u' = 1/x
Derivative of ln x
Write v' = 1 next to u' = 1/x.
→ Write v = x next to u = ln x.
→ u' = 1/x
Derivative of ln x
Write v' = 1 next to u' = 1/x.
→ Write v = x next to u = ln x.
[3]
uv: (ln x)⋅x
u'v: (1/x)⋅x
u'v: (1/x)⋅x
Close
Example
∫(ln x)2 dx
Solution ∫(ln x)2 dx
u v'
= ∫(ln x)2⋅1 dx
u = (ln x)2 v = x
u' = 2(ln x)1⋅1x
= (2 ln x)⋅1x v' = 1 - [1]
= (ln x)2⋅x - ∫(2 ln x)⋅1x⋅x dx - [2]
= x (ln x)2 - 2∫ln x dx
= x (ln x)2 - 2(x ln x - x + C)
= x (ln x)2 - 2x ln x + 2x + C
u v'
= ∫(ln x)2⋅1 dx
u = (ln x)2 v = x
u' = 2(ln x)1⋅1x
= (2 ln x)⋅1x v' = 1 - [1]
= (ln x)2⋅x - ∫(2 ln x)⋅1x⋅x dx - [2]
= x (ln x)2 - 2∫ln x dx
= x (ln x)2 - 2(x ln x - x + C)
= x (ln x)2 - 2x ln x + 2x + C
[1]
Write u = (ln x)2.
→ u' = 2(ln x)(1/x)
Derivative of g(f(x))
Write v' = 1 next to u' = (2 ln x)⋅(1/x).
→ Write v = x next to u = (ln x)2.
→ u' = 2(ln x)(1/x)
Derivative of g(f(x))
Write v' = 1 next to u' = (2 ln x)⋅(1/x).
→ Write v = x next to u = (ln x)2.
[2]
uv: (ln x)2⋅x
u'v: 2(ln x)(1/x)⋅x
u'v: 2(ln x)(1/x)⋅x
Close