Silly question about integration of a fraction

Question

I was solving a problem with partial fractions but I am going to skip all of that.

Lets say we are given $f(x)= \frac{\frac{-9}{77}}{2x+5}$ and we want to find the indefinite integral:

$$\int \frac{\frac{-9}{77}}{2x+5}\,dx$$ Letting:

$u=2x+5$ $du=2$

then

$\int \frac{\frac{-9}{77}}{2x+5}\,dx = \frac{-9}{77}\int\frac{1}{u}du= \frac{-18}{77}ln \vert 2x+5 \vert$

Am I missing some basic algebreic principle? Because I know that the answer should come out to be $\frac{-9}{144} ln \vert 2x+5 \vert$ Why should $du$ be multiplied into the denominator and not the numerator?

Answer 1

First, when you write $du=2$, this is inaccurate, it is $du=2dx$.

In light of this, you realise that your integral was missing a $dx$ from the beginning.

When you put back the missing $dx$ and then replace it with $\frac12 du$ as it should be, you get the right answer.

Answer 2

You can also do simply this $$I=\int \frac{\frac{-9}{77}}{2x+5}\,dx$$ $$I=\frac{-9}{77}\int \frac {dx} {2x+5}$$ you can factorize $1/2$ $$I=\frac{-9}{2 \times 77}\int \frac {dx} {x+5/2}$$ $$I=\frac{-9}{2 \times 77}\ln[{x+5/2}|+K$$ $$I=\frac{-9}{2 \times 77}\ln[{2x+5}|+C$$

Answer 3

You can do

$2x+5=u \longrightarrow 2dx=du \longrightarrow dx=\frac{du}{2}$

then

$\int(\frac{\frac{-9}{77}}{2x+5}) dx = \int(\frac{\frac{-9}{77}}{u} \frac{du}{2} = \frac{-9}{77} \int(\frac{1}{u}) \frac{du}{2} = \frac{-9}{144} \int(\frac{du}{u}) = \frac{-9}{144} \ln |2x+5| $