# Simplifying changes integration

1. Jul 4, 2011

### gersetaffe

Hi,

Wanted to know if anyone could explain why if you simplify an expression into a different equivalent form, the integrations are different depending on which form you use.

For example:
$\frac{1}{\frac{5x}{7}+3}$ = $\frac{1}{\frac{5}{7}(x+4.2)}$

$\int$$\frac{1}{\frac{5x}{7}+3}$dx = $\frac{7}{5}$ln($\frac{5x}{7}$+3)
while
$\int$$\frac{1}{\frac{5}{7}(x+4.2)}$dx = $\frac{7}{5}$ln(x+4.2)

The two integrations are not equal despite having integrated two equivalent expressions. The issue is if I had to integrate $\frac{1}{\frac{5x}{7}+3}$ I would simplify it to
$\frac{1}{\frac{5}{7}(x+4.2)}$ which gives a different integration than the original expression.

Thanks for any input

2. Jul 4, 2011

### JJacquelin

In the given example, both results are false because the constants have been forgotten.
If the constants C1 for the first integral and C2 for the second integral were presents, both results are exact and identical.
(7/5)*ln((5/7)x+3) +C1 = (7/5)*ln(x+4.2) + C2
C2 = (7/5)*ln(5/7) +C1

3. Jul 4, 2011

### micromass

Hi gersetaffe!

You shouldn't forget the constants when integrating!! Thus

$$\int\frac{1}{\frac{5}{7}x+3}dx=\frac{7}{5}ln(\frac{5}{7}x+3)+C$$

and

$$\int \frac{1}{\frac{5}{7}(x+4.2)}dx=\frac{7}{5}ln(x+4.2)+C'$$

these two solutions are the same because the integration constants are different:

$$\frac{7}{5}ln(\frac{5}{7}x+3)=\frac{7}{5}ln(5/7)+\frac{7}{5}ln(x+4.2)$$

Thus you see that the two solutions are equal up to a contant, namely, the constant $\frac{7}{5}ln(5/7)$.

So

$$\frac{7}{5}ln(\frac{5}{7}x+3)+C=\frac{7}{5}ln(x+4.2)+C'$$

with $C'=C+\frac{7}{5}ln(5/7)$.

4. Jul 4, 2011

### gersetaffe

My high school calculus teacher always told us to remember the constants. Should have listened haha ... thanks JJacquelin and micromass