Applying partial fractions in ∫ #10;1( x + 2)( x + 3)dx . #10; #10; 1( x + 2)( x + 3) #10;= Ax + 2 + Bx + 3 #10; #10; 1 = A( x + 3) + ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Integration by Partial Fractions

∫1x+2x+3dx

Question

$\int \frac{1}{(x + 2) (x + 3)} d x$

Open in App

Solution

Applying partial fractions in $\int \frac{1}{(x + 2) (x + 3)} d x$ .

$\frac{1}{(x + 2) (x + 3)} = \frac{A}{x + 2} + \frac{B}{x + 3}$

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

Put $x = - 3$ ,

$1 = - B$

$B = - 1$

Put $x = - 2$ ,

$1 = A$

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

$\int \frac{1}{(x + 2) (x + 3)} d x = \int \frac{1}{x + 2} d x - \int \frac{1}{x + 3} d x$

$= log | x + 2 | - log | x + 3 | + c$

Suggest Corrections

Similar questions

\int \frac{(2 x + 1)}{(x + 2) (x - 3)} d x

$\int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x$

\int \sqrt{\frac{x - 1}{2 x - 3}} d x =

\int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x

\int \frac{x + 1}{(x + 2) (x + 3)} d x

Join BYJU'S Learning Program