Question
Find $\int \frac{d x}{(x+1)(x+2)}$
✓
Answer
The integral is a proper rational function.
Therefore, by using the form of partial fraction, we write
$\frac{1}{(x+1)(x+2)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+2}$ ......(i)
where, real numbers A and B are to be determined suitably. This gives
1 = A (x + 2) + B (x + 1).
Equating the coefficients of x and the constant term, we get
A + B = 0 and 2A + B = 1
Solving these equations, we get A = 1 and B = -1.
Thus, the integral is given by
$\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}+\frac{-1}{x+2}$
Therefore, $\int \frac{d x}{(x+1)(x+2)}=\int \frac{d x}{x+1}-\int \frac{d x}{x+2}$
= log |x + 1| - log |x + 2| + C
= $\log \left|\frac{x+1}{x+2}\right|+C$
Therefore, by using the form of partial fraction, we write
$\frac{1}{(x+1)(x+2)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+2}$ ......(i)
where, real numbers A and B are to be determined suitably. This gives
1 = A (x + 2) + B (x + 1).
Equating the coefficients of x and the constant term, we get
A + B = 0 and 2A + B = 1
Solving these equations, we get A = 1 and B = -1.
Thus, the integral is given by
$\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}+\frac{-1}{x+2}$
Therefore, $\int \frac{d x}{(x+1)(x+2)}=\int \frac{d x}{x+1}-\int \frac{d x}{x+2}$
= log |x + 1| - log |x + 2| + C
= $\log \left|\frac{x+1}{x+2}\right|+C$
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