(i) `int cf(x) dx = c int f(x )dx`, where `c` is any constant.

(ii) `int [f_1 (x) pm f_2(x) ± f_2 (x) ± f_3 ±f_4(x) ± ... ]dx ,= int f_(x)dx ± f_2(x)dx ± int f_3(x)dx ± int f_4(x)dx ± ...`

(iii) If `int f(x)dx = g(x) +C` , then `int f(ax ± b) dx = 1/a g (ax± b)+ C`

Algebraic Formulae

(i) `int x^n dx =(x^(n+1))/(n+1) + C , n ne -1`

(ii) `int(a x +b)^n dx = 1/a * ((ax+b)^(n+1))/(n+1) + C , n ne -1`

(iii) `int 1/x dx = log |x| + C`

(iv) `int 1/(ax+b) dx = 1/a (log |ax+b|) + C`

(v) `int 1/(a^2 -x^2) dx = 1/(2a) log |(a+x)/(a-x)| + C`

(vi) `int 1/(x^2-a^2) dx = 1/(2a) log |(x-a)/(x+a)|+ C`

(vii) `int 1/(sqrt(a^2-x^2)) dx = sin^(-1) \ x/a + C`

(viii) `int (-1)/(sqrt(a^2-x^2)) dx = cos^(-1) \ x/a + C`

(ix) `int 1/(a^2+x^2) dx = 1/a tan^(-1) \ x/a + C`

(x) `int (-1)/(a^2+x^2) dx = 1/a cot^(-1) \ x/a + C`

(xi) `int 1/(x sqrt(x^2-a^2)) dx = 1/a sec^(-1) \ (x/a) + C`

(xii) `int (-1)/(x sqrt(x^2 -a^2)) dx = 1/a cosec^(-1) \ (x/a) + C`

(xiii) `int 1/(sqrt(x^2-a^2)) dx =log |x + sqrt(x^2-a^2)|+ C`

(xiv) `int 1/sqrt(x^2+a^2) dx = log|x + sqrt(x^2-a^2)|+ C`

(xv) `int(sqrt a^2 -x^2) dx = 1/2 x sqrt(a^2-x^2) + 1/2 a^2 sin^(-1) \ (x/a) + C`

(xvi) `int sqrt(x^2-a^2) dx =1/2 x sqrt(x^2-a^2)-1/2 a^2 log|x + sqrt(x^2-a^2)|+ C`

(xvii) `int sqrt(x^2+a^2) dx =1/2 x sqrt(x^2+a^2) + 1/2 log|x + sqrt(x^2 +a^2)|+ C`

(i) `int sin x dx = - cosx + C`

(ii) `int cos x = sinx + C`

(iii) `int tan x dx =-log |cosx |+ C =log | secx I+ C`

(iv) `int cot x dx = log | sinx |+ C =- log | cosec x|+ C`

(v) `int secx dx =log | secx+tanx |+C=log | tan (pi/4+x/2) |+C`

(vi) `int cosecx dx = log | cosecx -cot x | +C = log | tan \ x/2 | +C`

(vii) `int sec^2x dx = tanx + C`

(viii) `int cosec^2x dx =- cotx + C`

(ix) `int sec xtanx dx =sec x + C`

(x) `int cosec x cotx dx =- cosecx + C`

(i) `int e^x dx = e^x + C`

(ii) `int e^((ax+b)) dx =1/a -e^((ax +b))+ C`

(iii) `int a^x dx = (a^x)/(log_e a) + C`

(iv) `int a^((bx+c))dx = 1/b * (a^(bx+c))/(log_e a) + C`

Mainly, there are three methods of integration which are given below:

(i) Direct Substitution :

lf `int f(x)dx = g(x) + C`, then to solve `I= int f{h(x)} *h'(x) dx`

we put `h(x)=t => h'(x)dx=dt`

So, `I= int f(t)dt=g(t)+C=> g{h(x)}+C`

Some Standard Integral

`=> int f(x)^n f'(x)dx = ([f(x)]^(n+1))/(n+1)+ C`

`=> int (f'(x))/(f(x)) dx = log |f(x)| + C`

`=> int (f' (x))/(sqrt(f(x))) dx = 2 sqrt (f(x)) + C`

(ii) Standard Substitution :

Some standard integrands or a part of it have standard substitution. Some standard substitutions are given below :






(iii) Indirect Substitution :

If integrand `f(x)` can be rewritten as product of two functions, i.e. `f(x)= f_1(x)* f_2(x)`, where `f_2(x)` is a function of integral of `f_1 (x )`, then put integral of `f_1 (x) = t`.

If integrand `f(x)` can be expressed as product of two functions, i.e. `f(x) = f_1 (x) * f_ 2(x)`, then we use the following
formula:

`int f_1(x)*f_2(x)dx =f_1 (x) int f_2(x)dx- int {f_1'(x)[int f_2(x)dx]} dx`

where, `f_1 (x)` and `f_2(x)` are known as first and second functions respectively.

i.e, The integral of the product of two functions = (First function) x (Integral of the second function)
- Integral of {(Differentiation of first function) x (Integral of second function)}

If integrand is a rational function, i.e. of the form `(f(x))/(g(x))` where f(x) and g(x) are the polynomial functions of x and
`g(x) ne 0`, then use method of partial fractions, if degree of `f(x)` is less than the degree of `g(x)`. Otherwise, first
reduced to the proper rational function by long division process. The following table indicates the types of simpler
partial fractions that are associated with various kinds of rational functions.

Here A, B and C are the real constants and these can be determined by reducing both sides of the equation as identity in polynomial form and by comparing the coefficients of like powers.

(i) `int f(x) dx = int f(x) * 1 dx`

Now, integrate taking `f(x)` as a first and 1 as a second function.

(ii) `int (f(x))/([g(x)]^n) dx = int (f(x))/(g'(x)) * (g'(x))/([g(x)]^n) dx`

Now, integrate by taking `(f(x))/(g'(x))` as first function and `(g'(x))/([g(x)]^n)` as a second function

Here, ` f(x)` and `g(x)` both are logarithmic functions.

Integration of the Form `int e^x {g(x)+g'(x) } dx`:

If integrand is of the form `e^x {g(x)+ g'(x)}`, then `int e^x *[g(x) + g'(x)] dx = e^x g(x)+C`

Integration of the Form `int {g(x)+ xg'(x) } dx`:

If integrand is of the form then `int [g(x) + xg'(x)] dx = x g(x)+C`

`Type-1 :` `int dx /( (x-alpha) sqrt ((x-alpha)(beta-x))) ` `(beta > alpha)` (start : `x = alpha cos^2 theta + beta sin^2 theta`)

`Type-2:` `int dx /((ax+b) sqrt(px +q) )` ; e.g., `int dx /((2x +1) * sqrt (4x+3))`

Put `px+q=t^2`

`Type -3:` `int dx /((ax+b) sqrt (px^2 +qx +r))` ; e.g. `int dx/((x+1)*sqrt(1+x-x^2))`

Put `ax+b =1/t`

`Type-4:` `int dx /( (ax^2 +bx +c) sqrt (px+q))` ; put `px +q=t^2`

e.g. `int dx/((x^2 +5x +2)sqrt (x-2))` this reduces to `2 int dt /(t^4 +9t^2+16)`

`Type - 5:` `int dx /( (ax^2 +bx +c) sqrt (px^2 +qx +r))`

`text(Case-I :)` When `(ax^2 + bx +c)` breaks up into two linear factors, e.g.

`I = int dx/ ((x^2-x-2) sqrt (x^2 +x +1)) ` then

`= int ( (A/(x-2))+(B/(x+1))) (1/ (sqrt (x^2 +x+1))) dx = A int dx /( underbrace((x-2) sqrt(x^2 +x+1))_(text(put) x-2 =1/t) )+ B int dx /( underbrace((x+1) sqrt(x^2 +x+1))_(text(put) x+1 =1/t) )`

`text(Case-II:)` If `ax^2 + bx + c` is a perfect square say `(lx + m)^2` then put `lx + m = 1/t`

`text(Case-III:)` If `b = 0; q = 0` e.g. `int dx /((ax^2 +b) sqrt (px^2 +r))` then put `x=1/t` or the trigonometric substitution are also helpful.
e.g. `int dx /((x^2 +4)sqrt (4x^2+1))`

Type - 1 : `int (dx )/(a + b sin^2 x) / int (dx )/(a + b cos^2 x) / int (dx )/(a sin^2 x + b cos^2 x + csin x cos x ) / int (dx )/(a cos x + b sin x)^2`

Multiply `sec^2x` in `N^r` and `D^r` and solve

Type - 2 : `int (dx )/(a + b sin x) / int (dx )/(a + b cos x) / int (dx )/(a + b sin^2 x) / int (dx )/(a + b sin x + c cos x)`

Convert `sin x` and `cos x` into their corresponding tangent to half the angles and

`sinx = 2tan \ x/2 / 1 + tan^2 \ x/2`

`cos = 1 - tan^2 \ x/2 / 1 + tan^2 \ x/2`

and put ` tan \ x/2 = t` solve

Type - 3 : `int ( a sin x + b cos x + c)/(l sin x + m cos x + n ) dx ; N^r = A(D^r) + B (d/(dx) D^r) + C`

Type - 4 : `int ( x^2 + 1)/(x^4 + Kx^2 + 1) dx ` or ` int ( x^2 - 1)/(x^4 + Kx^2 + 1) dx `

Divide `N^r` and ` x^2` and take suitable substitution