Evaluate the following integral:
Denominator is factorised, so let separate the fraction through partial fraction, hence let
We need to solve for A, B, C and D. We will equate the like terms we get,
C = 0………….(iii)
A + D = 1⇒ A = 1 – D………(iv)
2A + B + C = 1
⇒ 2(1 – D) + B + 0 = 1 (from equation (iii) and (iv))
⇒ B = 2D – 1……….(v)
2B + D = 1
⇒ 2(2D – 1) + D = 1 (from equation (v), we get
⇒ 4D – 2 + D = 1
⇒ 5D = 3
……………(vi)
Equation (vi) in (v) and (iv), we get
We put the values of A, B, C, and D values back into our partial fractions in equation (i) and replace this as the integrand. We get
Split up the integral,
Let substitute
u = x2 + 1 ⇒ du = 2xdx,
y = x + 2 ⇒ dy = dx, so the above equation becomes,
On integrating we get
(the standard integral of )
Substituting back, we get
Note: the absolute value signs account for the domain of the natural log function (x>0).
Hence,