3
$\begingroup$

Integrate $\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$

My Attempt: $$\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$$ $$\int \dfrac {1}{x+2} \textrm {dx} . \int \dfrac {1}{x+3} \textrm {dx}$$ $$\dfrac {\textrm {log (x+2)}}{1} . \dfrac {\textrm {log (x+3)}}{1} + C$$ $$\textrm {log} (x+2) . \textrm {log} (x+3) + C$$

Is this correct? Or, How do I proceed the other way?

$\endgroup$
  • 4
    $\begingroup$ nooooo, you can't just do it $\endgroup$ – Saketh Malyala May 1 '17 at 5:00
  • 3
    $\begingroup$ $$\int(uv)dx$$ in general $$\ne\int u\ dx\cdot\int v\ dx$$ Use $$1=x+3-(x+2)$$ $\endgroup$ – lab bhattacharjee May 1 '17 at 5:00
  • $\begingroup$ You can't just split multiplication inside the integral to multiplication outside the integral. Have you tried a partial fraction decomposition? $\endgroup$ – DMcMor May 1 '17 at 5:01
  • $\begingroup$ @labbhattacharjee, Could you please elaborate your hint? $\endgroup$ – pi-π May 1 '17 at 5:04
  • $\begingroup$ you cannot split a product into multiple integrals. you must simplify the algebraic expression into integrable pieces. one way of doing this is using the method of partial fractions to simplify the quadratic denominator into a sum of linear denominator - fractions $\endgroup$ – Saketh Malyala May 1 '17 at 5:05
3
$\begingroup$

Method to do -

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

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

Case 1 -

When $x+3=0$

$x=-3$

Put $x=-3$

$1 = -B$

$B = -1$

Case 2 -

When $x+2=0$

$x=-2$

Put $x=-2$

$1 = A$

$A = 1$

Now your integral becomes,

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

$\endgroup$
  • $\begingroup$ Which method did you use? Partial Fraction? $\endgroup$ – pi-π May 1 '17 at 5:18
  • $\begingroup$ Yes partial fraction. $\endgroup$ – Kanwaljit Singh May 1 '17 at 5:35
10
$\begingroup$

Nope! That's not how integration works.

You want to split the denominator using partial fractions.

$\displaystyle\int \frac{1}{(x+3)(x+2)}\,dx =\int \left(\frac{1}{x+2}-\frac{1}{x+3}\right)\,dx=\ln|x+2|-\ln|x+3|+C=\boxed{\ln \left|\frac{x+2}{x+3}\right|+C}$

$\endgroup$
0
$\begingroup$

The method you need to use is Partial Fractions.

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

The LCD is (x+2)(x+3.)

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

Then you set both of the numerators equal to each other:

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

$$\\$$ When does x+3 = 0?

x= -3

Plug -3 into x:

1 = $\require{cancel} \cancel{A(-3+3)}$ + B(-3+2)

$\implies$ 1 = -1B

$\implies$ -1 = B

$$\\$$

When does x+2 = 0?

x= -2

Plug -2 into x (and -1 into B because we solved for B above):

1 = A(-2+3) + $\require{cancel} \cancel{-1(-2+2)}$

$\implies$ 1 = 1A

$\implies$ 1 = A

$$\\$$ Now plug in A= 1 and B= -1 into: $\frac{A}{x+2}+\frac{B}{x+3}$. Thus your integral is $\int (\frac{1}{x+2}+\frac{-1}{x+3})\,dx$.

When you integrate you get: $\boxed{\ln \left|{x+2}| - \ln|x+3\right|+C}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.