Let I = ∫▒e^x/(e^4x+e^2x+1) dx, J = ∫▒e^(-x)/(e^(-4x)+e^(-2x)+1) dx Then the value of J-I equals:

Let I = ∫▒e^x/(e^4x+e^2x+1) dx, J = ∫▒e^(-x)/(e^(-4x)+e^(-2x)+1) dx Then, for any arbitrary constant C, the value of J-I equals:
a) 1/2 log⁡((e^4x-e^2x+1)/(e^4x+e^2x+1)) + C b) 1/2 log⁡((e^2x+e^x+1)/(e^2x-e^x+1)) + C
c) 1/2 log⁡((e^2x-e^x+1)/(e^2x+e^x+1)) + C d) 1/2 log⁡((e^4x+e^2x+1)/(e^4x-e^2x+1)) + C

Integral Types
The integral calculus is of the two forms, namely

(i) Indefinite Integral

(ii) Definite Integral

In an indefinite integral, the range of the function is not defined, thus the value of the function obtain is followed by a constant value ‘c.’

Whereas in a definite integral, the range of the function is well defined, thus it gives a well-defined function.

The integration is denoted by (∫).

What are Indefinite Integrals?

Properties of Indefinite Integrals
Integrals follow certain properties as follows:

∫[f(x)+g(x)dx]=∫f(x)dx+∫g(x)dx
For any given real number r, ∫rf(x)dx]=r∫f(x)dx
If f1, f2, f3 … are the functions and r1,r2,r3… are the designated real numbers. Then,
∫[r1f1(x)+r2f2(x)+…+rnfn(x)]dx=r1∫f1(x)dx+r2∫f2(x)dx+…+rn∫fn(x)dx
Integration using Partial Fractions
The rational function property can be seen in fractions and is followed by the ratio of two polynomials such as R(x)/S(x) where x and S(x) ≠ 0. If R(x) degree is higher than S(x), then we should divide R(x) by S(x) so that R(x)/S(x)=T(x)+ R1(x)/S(x) , where T(x) is a degree x polynomial and the degree of R1(x) is less than that of S(x). T(x) can be easily integrated.

Integration using Substitution
In this method, we change the variable of another variable in order to reduce the integral into fundamental integrals. We can get some standard integrals as the following.

∫tanxdx=log|secx|+C
∫cotxdx=log|sinx|+C
∫secxdx=log|secx+tanx|+C
∫cosecxdx=log|cosecx–cotx|+C

Example 1: Find the integral of the function: ∫30x2dx

Solution:

Given ∫30x2dx

= (x33)30

=(333)–(033)

= 9

Example 2: Find the integral of the function: ∫x2 dx

Solution:

Given ∫x2 dx

= (x3/3) + C.
If the integrand is not a derivative of a known function, the integral may be evaluated with the help of any of the following three rules:

1) Integration by substitution or by change of the independent variable.

2) Integration by parts

3) Integration by partial fractions

Some indefinite integrals which can be evaluated by direct substitutions:
1) If integral is of the form ∫ f(g(x)) g'(x) dx, then put g(x) = t, provided ∫ f(t) exists.

2) ∫ f'(x)/f(x) dx = ln |f (x)| + c, By putting f (x) = t = f' (x) dx = dt

= ∫ dt/t = ln |t| + c = ln |f (x)| + c.

3) ∫ f'(x)√f(x) dx = 2 √f(x)+c, Put f (x) = t

Then ∫ dt/√t = 2√t + c = 2√f(x) + c.

Some standard substitutions:
1) For terms of the form x2 + a2 or √x2 + a2, put x = a tanθ or a cotθ

2) For terms of the form x2 - a2 or √x2 – a2 , put x = a sec θ or a cosecθ

3) For terms of the form a2 - x2 or √x2 + a2, put x = a sin θ or a cosθ

4) If both √a+x, √a–x, are present, then put x = a cos θ.

5) For the form √(x–a)(b–x), put x = a cos2θ + b sin2θ

6) For the type (√x2+a2±x)n or (x±√x2–a2)n, put the expression within the bracket = t.

7) For 1/(x+a)n1 (x+b)n2, where n1,n2 ∈ N , again put (x + a) = t (x + b)

If the integrand is of the form f(x)g(x), where g(x) is a function of the integral of f(x), then put integral of f(x) = t.

The integral of product of two functions of x is evaluated with the help of integration by parts. Let u and v be two functions of x, then ∫uv dx = u∫v dx - ∫[du/dx ∫v dx]dx

While carrying out integration by parts, whether a function is u or v should be decided according to ILATEmethod of integration (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent).

If both the functions are directly integrable then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable.

If in the product of the two functions, one of the functions is not directly integrable like lnx, sin-1x, cos-1x, tan-1x etc. then we take it as the first function and the remaining function is taken as the second function.

If there is no second function available, then unity is taken as the second function e.g. in the integration of∫tan-1x dx, tan-1x is taken as the first function and 1 as the second function.

In the integral ∫g(x)exdx, if g(x) can be expressed as g(x) = f(x) + f'(x) then ∫g(x)ex dx=∫ ex [f(x) + f'(x)] dx = exf(x) + c

To write P(x)/ Q(x) in partial fractions, write Q(x) in the form Q(x) = (x – a)k ... (x2 + αx + β)r ... where binomials are different, and then set