Question

int ((3)/(4-7x)) dx

Work would be great, please and ty

Answer 1

your missing a -7

Answer 2

\[-\frac{3}{7}\int\frac{-7}{4-7x}dx \]

Answer 3

What do you mean i was missing a -7?

Answer 4

i mean that if you had a -7 you could simply ln it up

Answer 5

we can always apply a useful form of "1" to borrow a constant from without changing the value of anything

Answer 6

\[\frac{-7}{-7}\int\frac{3}{4-7x}dx\to \ ` -\frac{3}{7}\int\frac{-7}{4-7x}dx\]

Answer 7

Move the 3 outside of the integral (it's constant) and do the substitution
\[ u=4-7x \\
du = -7dx \]
or
\[ dx = \frac{du}{-7} \]
We now get:
\[ 3\int{\frac{1}{u}\frac{du}{-7}} = -\frac{3}{7}\int{\frac{du}{u}} = -\frac{3}{7}\ln{u} \]
By putting our substitution back we get:
\[ \int{\frac{3}{4-7x}dx} = -\frac{3}{7}\ln|4-7x| \]

Answer 8

Great explanation, thank you!

Answer 9

No problem. Also, if you didn't already know, the absolute value inside the ln is required because it isn't defined for negative values.