PSI Mathematics.

Integration Formulas of Inverse Trigonometric functions:
∫1/√(1 – x2).dx = sin-1x + C
∫ /1(1 – x2).dx = -cos-1x + C
∫1/(1 + x2).dx = tan-1x + C
∫ 1/(1 +x2 ).dx = -cot-1x + C
∫ 1/x√(x2 – 1).dx = -cosec-1 x + C
∫ 1/x√(x2 – 1).dx = sec-1x + C
Advanced Integration Formulas
∫ 1/(a2 – x2).dx =1/2a.log|(a + x)(a – x)| + C
∫1/(x2 – a2).dx = 1/2a.log|(x – a)(x + a| + C
∫1/(x2 + a2).dx = 1/a.tan-1x/a + C
∫1/√(x2 – a2)dx = log|x +√(x2 – a2)| + C
∫1/√(a2 – x2).dx = sin-1 x/a + C
∫ √(x2 – a2).dx =1/2.x.√(x2 – a2)-a2/2 log|x + √(x2 – a2)| + C
∫√(a2 – x2).dx = 1/2.x.√(a2 – x2).dx + a2/2.sin-1 x/a + C
∫1/√(x2 + a2 ).dx = log|x + √(x2 + a2)| + C
∫ √(x2 + a2 ).dx =1/2.x.√(x2 + a2 )+ a2/2 . log|x + √(x2 + a2 )| + C
Different Integration Formulas
T3 types of integration methods are generally used: Integration by parts formula, Integration by Substitution formula and Integration by partial fractions formula. Let us look at each of these integration formulas one by one.

Integration by Parts Formula
When any given function is a product of two different functions, the integration by parts formula or partial integration can be applied to evaluate the integral. The integration formula using partial integration methos is as follows:

∫ f(x).g(x) = f(x).∫g(x).dx -∫(∫g(x).dx.f'(x)).dx + c

For instance: ∫ xex dx is of the form ∫ f(x).g(x). Therefore, we must apply the appropriate integration formula and evaluate the integral accordingly.

f(x) = x and g(x) = ex

Thus ∫ xex dx = x∫ex .dx – ∫( ∫ex .dx. x). dx+ c

= xex – ex + c

Integration by Substitution Formula
If a given function is a function of another function, we can apply the integration formula for substitution to solve that integral. For instance, if
I = ∫ f(x) dx,
where
x = g(t) so that dx/dt = g'(t), then we write dx = g'(t)
Take for instance
I = ∫ f(x) dx = ∫ f(g(t)) g'(t) dt
For example: Consider ∫ (3x +2)4 dx
The integration formula of substitution is given as follows.
Take u = (3x+2). ⇒ du = 3. dx
Thus ∫ (3x +2)4 dx =1/3. ∫(u)4. du
= 1/3. u5 /5 = u5 /15
= (3x+2)5 /15