The value of \(\int \frac {x^2}{x^2 - 3x + 2}dx\) will be ___________, where c is an arbitrary constant.

Question

Question

This question was previously asked in

AAI (ATC) Junior Executive Official Paper (Held On: 30 Nov 2018 Shift 2)

Answer 1

Concept:

∫ 1 dx = x + constant

\(\int\frac{1}{x}dx=log\ x+constant\)

Calculation:

Given:

Let, \(I=∫ \frac {x^2}{x^2 - 3x + 2}dx\)

\(I= ∫ (1+\frac{3x-2}{x^2-3x+2})dx\)

\(I= ∫ 1dx+∫\frac{3x+2}{x^2-3x-2}dx\).....(i)

\(\frac{3x-2}{x^2-3x+2}=\frac{3x-2}{(x-2)(x-1)}=\frac{A}{(x-2)}+\frac{B}{(x-1)}\) ....(ii)

(3x - 2) = A (x - 1) + B (x - 2)

for x = 1

(3 (1) - 2) = B (1 - 2)

B = -1

for x = 2

(3 (2) - 2) = A (2 - 1)

A = 4

from equation (ii)

\(\frac{3x-2}{x^2-3x+2}=\frac{4}{(x-2)}-\frac{1}{(x-1)}\)

Now from equation (i)

\(I= ∫ 1dx+∫\frac{4}{(x-2)}dx-∫\frac{1}{(x-1)}dx\)

x - log |x - 1| + 4 log |x - 2| + c

where c is an arbitrary constant.