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 Math Help > Calculus > Integral > Table Of Integrals

### Tables of Integrals

On this page you will find a list of many of the most common integrals, organized by how "hard" they are.  "Hard" means it gets "messier" when you integrate it, and "easy" means it doesn't.  Moreover, the "hardest" integrals often have "easy" derivatives, and vice-versa.  So when you do integration by parts, (∫u dv = uv - ∫v du) you should pick dv as the easier, and u as the harder of the two functions to integrate, because you'll have to integrate dv and differentiate u.  You can remember the order of the categories because they spell "ILATE".  For dv, pick the latest one you can in that list.

 Function Category (click category to jump to that table) Comments I: inverse trigonometric functions: arctan(x), arcsec(x), etc. L: logarithmic functions: ln(x), log2(x), etc. A: algebraic functions: x2, 3x50, etc. T: trigonometric functions: sin(x), tan(x), etc. E: exponential functions: ex, 13x, etc. "Other"

I try to list the standard "integration by parts" integrals ( ∫u dv ) together with the part, dv, i.e. latest in the "ILATE" list.  So you'll find x cos(x) together with cos(x), for example, because x is algebraic (A), and cos(x) is trigonometric (T), which comes later in "ILATE".

Constants: I don't show the constant of integration, and in fact, the integrals shown may differ by a constant from what you would expect to see, and this is not wrong.  For example, the integral of x/(ax+b) is shown as (1/a2)(ax + b - b ln|ax + b|), because it's "nice" to see every "x" as part of "ax+b".  This, despite the method described giving you (1/a2)(ax - b ln|ax + b|).  These two integrals differ only by a constant, so they are both right.

Special treatment of the ln function: It's easy to forget that ln(u) = 2 ln(sqrt(u)), and sometimes the latter appears easier to integrate, for example, if u=x^2+a^2.  Similarly, if the ln function appears in the solution of an integral, it is important to remember that, for example, ln(u) and ln(3u) differ by a constant.

### I: inverse trigonometric functions: arctan(x), arcsec(x), etc.

 Function Indefinite Integral Comments arcsin x x arcsin x + sqrt(1-x2) arccos x x arccos x - sqrt(1-x2) arctan x x arctan x - ln(1+x2)/2

### L: logarithmic functions: ln(x), log2(x), etc.

 Function Indefinite Integral Comments ln(x) x ln(x) - x x ln(x) x2 (ln(x)/2 - 1/4) x2 ln(x) x3 (ln(x)/3 - 1/9) x3 ln(x) x4 (ln(x)/4 - 1/16) xn ln(x) x1+n (ln(x)/(n+1) - 1/(n+1)2) ln(x)/x (ln2(x))/2 1/(x ln(x)) ln(ln(x)) 1/(x ln(x) ln(ln(x))) ln(ln(ln(x))) 1/(x ln(x) ln(ln(x)) ln(ln(ln(x)))) ln(ln(ln(ln(x)))) etc. ln(x)/x2 -x-1 (1 + ln(x)) ln(x)/x3 -x-2 (1 + 2 ln(x))/4 ln(x)/xn -x1-n (1 + (n-1) ln(x))/(n-1)2

### A: algebraic functions: x2, 3x50, etc.

 Function Indefinite Integral Comments xn x(n+1)/(n+1) real n ≠ -1 x sqrt(x+a) (2/15)(3x-2a)(x+a)3/2 Integration by parts: ∫u dv = uv - ∫v du dv = sqrt(x+a) dx, v=(2/3)(x+a)3/2; u=x, du=dx; (2/3)(x)(x+a)3/2-(4/15)(x+a)5/2  ((2/3)(x)-(4/15)(x+a))(x+a)3/2  (2/15)(3x-2a)(x+a)3/2 x (x+a)k ((k+1)x-a)/((k+1)(k+2))(x+a)k+1 Integration by parts: ∫u dv = uv - ∫v du dv = (x+a)kdx, v=(1/(k+1))(x+a)k+1; u=x, du=dx; (1/(k+1))(x)(x+a)k+1-(1/((k+1)(k+2)))(x+a)k+2  (x/(k+1)-(x+a)/((k+1)(k+2)))(x+a)k+1   (x(k+2)-(x+a))/((k+1)(k+2))(x+a)k+1   ((k+1)x-a)/((k+1)(k+2))(x+a)k+1 (x+a)-1/2 (x+b)-3/2 2/(b-a) (x+a)1/2 (x+b)-1/2 a≠b;Proof: differentiate to get -(a-x)1/2/((b-a)(x+b)3/2) + 1/((b-a)(x+b)1/2(x+a)1/2)put this over a common denominator...(-(x+a)+(x+b)) / ((b-a)(x+a)1/2(x+b)3/2) (x+a)k-1 (x+b)-k-1 1/(k(b-a)) (x+a)k (x+b)-k a≠b, k nonzero;this generalizes the expression above,and is proved by differentiating the result..

Special case: 1/x, more generally, du/u:

 Function Indefinite Integral Comments 1/x ln |x| 1/(ax+b) (1/a) ln |ax+b| x/(ax+b) (1/a2)(ax + b - b ln|ax + b|) Use polynomial division to express x/(ax+b) as (1/a)(1-b/(ax+b)), and then integrate to get (1/a2)(ax - b ln|ax + b|), which is the result up to a constant. du/u ln |u| e.g.  ∫ sin(x)/cos(x) dx = -ln |cos(x)| 1/(ax+bx^k) ln(x)/a + ln(x)/(a(k-1)) -ln(ax+bx^k)/(a(k-1)) factorize the "x" out of the denominator, then use partial fractions:  ∫dx/(ax+bx^k) = ∫dx/(ax) + ∫dx/(ax(k-1))    - ∫(a+bkx^(k-1))dx/(a(k-1)(ax+bx^k))

### T: trigonometric functions: sin(x), tan(x), etc.

 Function Indefinite Integral Comments sin x -cos x cos x sin x sec2 x tan x tan2 x = sec2x - 1 tan(x) - x (sec x)(tan x) = sin x / cos2 x sec x csc2 x -cot x cot2 x = csc2x - 1 -x - cot x (csc x)(cot x) = cos x / sin2 x -csc x tan x -ln cos x cot x = 1/(tan x) ln sin x sec x = (sec x tan x + sec2 x)/(sec x + tan x) = 1/(cos^2(x/2) - sin^2(x/2)) = sec^2(x/2) / (1-tan^2(x/2)) ln(sec x + tan x), or 2 arctanh(tan(x/2) csc x = (-csc x cot x + csc2 x)/(csc x - cot x) = = (1/2)(csc(x/2) sec(x/2)) = (1/2)sec^2(x/2)/tan(x/2) ln(csc x - cot x), or ln(tan(x/2)) sin2 x = (1/2)(1-cos2x+sin2x) (1/2)(x - sin x cos x) = x/2 - sin(2x)/4 cos2 x = 1 - sin2x x/2 + sin(2x)/4 sin(ax)sin(bx) (1/2)(sin(a-b)x/(a-b) - sin(a+b)x/(a+b)) a ≠ ±b, Proof sin(ax)cos(bx) -(1/2)(cos(a-b)x/(a-b) + cos(a+b)x/(a+b)) a ≠ ±b, Proof cos(ax)cos(bx) (1/2)(sin(a-b)x/(a-b) + sin(a+b)x/(a+b)) a ≠ ±b, Proof x sin(ax) (sin(ax) - ax cos(ax))/a2 x cos(ax) (cos(ax) + ax sin(ax))/a2 sinh x cosh x cosh x sinh x tanh x ln cosh x

### E: exponential functions: ex, 13x, etc.

 Function Indefinite Integral Comments eax+b (1/a)(eax+b) cax+b (1/(a ln c))(cax+b) cax+b = (eln c)ax+b      = e(ln c)(ax+b) xeax ((ax-1)/a2)(eax) eaxsin(bx) (eax/(a2+b2)) (a sin(bx) - b cos(bx)) eaxcos(bx) (eax/(a2+b2)) (a cos(bx) + b sin(bx))

### "Other"

Functions involving (a2+x2) or (a2-x2) or its square root.
Trig substitution is a technique that helps with many of these.

 Function Indefinite Integral Comments 1/(a2+x2) (1/a) arctan (x/a) 1/(a2-x2) (1/a)(ln(1+x/a)-ln(1-x/a))/2 (1/a) arctanh (x/a) trig subst x/a=sin(θ) sqrt(a2-x2) (a2/2)arctan(x/sqrt(a2-x2)) + (x/2)sqrt(a2-x2) sqrt(x2-a2) -(a2/2)ln(x+sqrt(x2-a2)) + (x/2)sqrt(x2-a2) sqrt(a2+x2) (a2/2)ln(x+sqrt(a2+x2)) + (x/2)sqrt(a2+x2) 1/sqrt(a2-x2) arcsin (x/a) 0 ≤ x ≤ a 1/sqrt(x2-a2) ln(x + sqrt(x2-a2)) = arccosh (x/a) 0 ≤ a ≤ x; formulas differ by ln(a); trig subst x/a=sec(θ) 1/sqrt(x2+a2) ln(x + sqrt(x2+a2)) = arcsinh (x/a) these two formulas also differ by ln(a); trig subst x/a=tan(θ) x/sqrt(a2-x2) -sqrt(a2-x2) 0 ≤ x ≤ a x/sqrt(x2-a2) sqrt(x2-a2) 0 ≤ a ≤ x x/sqrt(x2+a2) sqrt(x2+a2) x sqrt(a2-x2) (-1/3)(a2-x2) sqrt(a2-x2) 0 ≤ x ≤ a x sqrt(x2-a2) (1/3)(x2-a2) sqrt(x2-a2) 0 ≤ a ≤ x x sqrt(x2+a2) (1/3)(x2+a2) sqrt(x2+a2)

. . . . . . write up the integral of (x+2x-1/2)1/2dx, so we make the substitution
x=u4/3, which means
dx = (4/3)u1/3du
Substituting, we get (4/3)(u4/3+2u-2/3)1/2u1/3du.
you can distribute the u1/3 into the square root by squaring it... so we get
(4/3)(u6/3+2u0/3)1/2du
(4/3)(u2+2)1/2du
which is solved using sqrt(a2+x2), above, and then substituting back . . . . . .

. . . . . . add this as a separate page, proving the integral of 1/sqrt(a^2+x^2)...
The derivative with respect to x of ln(u) is (1/u)(du/dx). So the derivative of ln(x + sqrt(1+x^2)) is (1 + x/sqrt(1+x^2))/(x + sqrt(1+x^2)), which simplifies to 1/sqrt(1 + x^2) using the fact that when a/b = c/d, it follows that a/b = (a+c)/(b+d) as well.

. . . . . . add the following factoid:

int ln(1+x^2)dx = int 2 ln(sqrt(1+x^2))dx
let x = tan u, so dx = sec^2 u du, and sqrt(1+x^2)=sec u, so
= int 2 (sec^2 u)(ln(sec u)) du
= (tan u) (ln(sec^2 u)) - 2(tan u + u) (by parts)
= (tan u) (ln(1+tan^2 u)) - 2(tan u + u)
= x (ln(1+x^2) - 2x + 2 arctan x

more generally, int ln(x^2+a^2) = x ln(x^2 + a^2) - 2x + 2a arctan(x/a)

 Function Indefinite Integral Comments 1/(1+sqrt(ax)) (2/a) (sqrt(ax) - ln(1+sqrt(ax))) Proof 1/(1+x²)² (1/2)(x/(1+x2) + arctan(x)) Proof 1/(x+sqrt(1-x²)) ½ arcsin(x) + ½ ln (x + sqrt(1-x²) ) 1/(1+sqrt(1-x²)) = (1-sqrt(1-x²))/x² arcsin(x) + (sqrt(1-x²)-1)/x 1/(1-sqrt(1-x²)) = (1+sqrt(1-x²))/x² -arcsin(x) + (sqrt(1-x²)+1)/x 1/(1+sqrt(x²-1)) = sqrt(x²-1)/(x²-2) - 1/(x²-2) (-1/sqrt(2)) arctan(x/sqrt(2x^2-2)) + ln(x+sqrt(x^2-1)) + (ln(x+sqrt(2))-ln(sqrt(2)-x))/(2sqrt(2)) . . . . . .  proof needed 1/(1-sqrt(x²-1)) = . . . . . . integral needed cos x / (sin x + cos x) ½ x + ½ ln (sin x + cos x) (a²+b²) / (a x + b sqrt(1-x²)) b arcsin(x) + a ln(ax + b sqrt(1-x²)) (a²+b²) cos x / (a sin x + b cos x) bx + a ln(a sin x + b cos x) Proof (a²+b²) / (a tan x + b) same as above! (a²+b²) sin x / (a sin x + b cos x) ax - b ln(a sin x + b cos x) Proof (a²+b²)/(a + b cot x) same as above! 4/(b-x4/b3) 2 arctan(x/b) + ln(b+x) - ln(b-x) 4x/(4b2+x4/b2) arctan((x-b)/b) - arctan((x+b)/b) = -arctan(2b2/x2) See sqrt(tan(x)) sqrt(tan(x)) (sqrt(2)/4) ln(tan(x)-sqrt(2tan(x))+1) + (sqrt(2)/2) atan(sqrt(2tan(x))-1) + (-sqrt(2)/4) ln(tan(x)+sqrt(2tan(x))+1) + (sqrt(2)/2) atan(sqrt(2tan(x))+1) Proof (1-sin x)/(1+sin x) = sec2x - 2sec x tan x + tan2x 2(tan(x/2)-1)/(tan(x/2)+1) - x = 2 tan(x) - 2 sec(x) - x Proof  Discussion (x2-1)/(x4+1) (1/sqrt(8))(ln(x2-sqrt(2)x+1)-ln(x2+sqrt(2)x+1)) Proof Factor x4+1 as (x2-sqrt(2)x+1)(x2+sqrt(2)x+1); partial fractions (x3n-1-xn-1)/(x4n+1) (1/(sqrt(2)n)) (ln(x2n-sqrt(2)xn+1)-ln(x2n+sqrt(2)xn-1)) With substitution y=xn; then this becomes (1/n)(y2-1)/(y4+1),because (1/n)dy = xn-1 dx, and so (1/n)y2 dy = x3n-1 dx

### Proofs of Selected Integrals

I = sin(ax) sin(bx) dx

From the identity cos(a) - cos(b) = -2 sin((a-b)/2) sin((a+b)/2), which is proved here, we get:

(1/2) (cos(px) - cos(qx)) = sin((q-p)x/2) sin((q+p)x/2)

Let a=(q-p)/2, and let b=(q+p)/2; then a+b=q and a-b=-p, and remembering cos(-x)=cos(x),

(1/2) (cos((a-b)x) - cos((a+b)x)) = sin(ax) sin(bx)

So the integral, above, is equal to

∫ sin(ax) sin(bx) dx =
(1/2) ∫cos((a-b)x) - cos((a+b)x)  =
(1/2)(sin(a-b)x/(a-b) - sin(a+b)x/(a+b)) + C

I = sin(ax) cos(bx) dx

From the identity sin(a) + sin(b) = 2 cos((a-b)/2) sin((a+b)/2), which is proved here, we get:

(1/2) (sin(qx) + sin(px)) = sin((p+q)x/2) cos((p-q)x/2)

Let a=(p+q)/2, and let b=(p-q)/2; then a+b=p and a-b=q, and therefore,

(1/2) (sin((a-b)x) + sin((a+b)x)) = sin(ax) cos(bx)

So the integral, above, is equal to

∫ sin(ax) cos(bx) dx =
(1/2) ∫ sin((a-b)x) + sin((a+b)x)  =
-(1/2)(cos(a-b)x/(a-b) + cos(a+b)x/(a+b)) + C

I = cos(ax) cos(bx) dx

From the identity cos(a) + cos(b) = 2 cos((a-b)/2) cos((a+b)/2), which is proved here, we get:

(1/2) (cos(px) + cos(qx)) = cos((p+q)x/2) cos((p-q)x/2)

Let a=(p+q)/2, and let b=(p-q)/2; then a+b=p and a-b=q, and so,

(1/2) (cos((a+b)x) + cos((a-b)x)) = cos(ax) cos(bx)

So the integral, above, is equal to

∫ cos(ax) cos(bx) dx =
(1/2) ∫ cos((a-b)x) + cos((a+b)x)  =
(1/2)(sin(a-b)x/(a-b) + sin(a+b)x/(a+b)) + C

(1/2)(sin(a-b)x/(a-b) + sin(a+b)x/(a+b))

I = (a²+b²) cos x dx / (a sin x + b cos x)

The approach is to break this down into the sum of two integrals, I1 and I2, and introduce a third integral, I3, such that I2+I3 and I1-I3 are both easy to do, and so the sum of these two integrals is I1+I2, and that's our answer.

Let I1 = a2 cos x dx/(a sin x + b cos x)

Let I2 = b2 cos x dx/(a sin x + b cos x)

Let I3 = ab sin x dx/(a sin x + b cos x)

Now,

I1 - I3 = a (a cos x - b sin x) dx/(a sin x + b cos x)
= a ln | a sin x + b cos x | + C

I2 + I3 = b (a sin x + b cos x) dx/(a sin x + b cos x)
= b x + D

Now add up these equations, and you get

I1 + I2 = I =  b x + a ln | a sin x + b cos x | + E

I = (a²+b²) sin x dx / (a sin x + b cos x)

The approach is to break this down into the sum of two integrals, I1 and I2, and introduce a third integral, I3, such that I1+I3 and I2-I3 are both easy to do, and so the sum of these two integrals is I1+I2, and that's our answer.

Let I1 = a2 sin x dx/(a sin x + b cos x)

Let I2 = b2 sin x dx/(a sin x + b cos x)

Let I3 = ab cos x dx/(a sin x + b cos x)

Now,

I2 - I3 = -b (a cos x - b sin x) dx/(a sin x + b cos x)
= -b ln | a sin x + b cos x | + C

I1 + I3 = a (a sin x + b cos x) dx/(a sin x + b cos x)
= a x + D

Now add up these equations, and you get

I1 + I2 = I = a x - b ln | a sin x + b cos x | + E

I = (1-sin x)/(1+sin x) dx

∫(1-sin x)/(1+sin x) dx
= ∫(1-sin x)2/(1-sin2x) dx
= ∫(1 - 2 sin x + sin2x)/cos2x dx
= ∫sec2x dx  - ∫2 sin x/cos2x dx + ∫tan2x dx
= ∫sec2x dx  - ∫2 sin x/cos2x dx + ∫(sec2x -1) dx
= tan x - 2 sec x + (tan(x)-x) + C
= 2 tan(x) - 2 sec(x) - x + C

I = 1/(1+x²)²

Rewrite as ∫(1/2)(1+x²)/(1+x²)² dx + ∫(1/2)(1-x²)/(1+x²)² dx
The first integral is ∫(1/2)(1/(1+x²)) dx, which is (1/2)arctan(x)
The second integral is ∫(1/2)(1-x²)(1+x²)-2 dx, and the integral of this is (1/2)(x)(1+x²)-1
So the answer is (1/2)(x/(1+x²) + arctan(x))

I = (x2-1)/(x4+1) dx

Factor x4+1 as (x2-sqrt(2)x+1)(x2+sqrt(2)x+1)
Then let x2-1 = (Ax+B)(x2+sqrt(2)x+1) + (Cx+D)(x2-sqrt(2)x+1)
= (A+C)x3 + (sqrt(2)A+B-sqrt(2)C+D)x2 + (A+sqrt(2)B+C-sqrt(2)D)x + (B+D), which by equating coefficients, gives us the following system of linear equations:

A + C = 0
sqrt(2)A + B - sqrt(2)C + D = 1
A + sqrt(2)B + C - sqrt(2)D = 0
B + D = -1

Solving this system gives us A=sqrt(2)/2, B=-1/2, C=-sqrt(2)/2, D=-1/2, so

∫(x2-1)/(x4+1) dx = (1/2)∫((sqrt(2)x-1)/(x^2-sqrt(2)x+1) + (-sqrt(2)x-1)/(x^2+sqrt(2)x+1)) dx

= (1/sqrt(8))(ln(x2-sqrt(2)x+1)-ln(x2+sqrt(2)x+1))

### Internet references

M.G. Worster, Practice Integrals

### Related pages in this website

Trig Equivalences, which proves these identities:
(1-sin 2x)/(1+sin 2x) = (1 - tan(x))2/(1 + tan(x))2
1-cos(2u) = 2sin²u
tan x+y = (tan x + tan y)/(1 - tan x tan y), and when y=-π/4, this becomes
tan x-π/4 = (tan x - 1)/(tan x + 1)

Weierstrass t-substitution is a clever trig substitution that lets you solve the kind of integrals that naturally come up in the polar form of certain functions. The substitution is t=tan(x/2).

The webmaster and author of this Math Help site is Graeme McRae.